All nilpotent matrices $2 times 2$ satisfies $A^{2}=0$ [duplicate]












4















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  • All nilpotent $2times 2$ matrices

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I have problems to show that if $A$ is a $2 times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.










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Dec 9 at 16:24


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  • Yes, it was a mistake
    – DIEGO R.
    Nov 26 at 15:03










  • You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
    – Sam Streeter
    Nov 26 at 15:12
















4















This question already has an answer here:




  • All nilpotent $2times 2$ matrices

    2 answers




I have problems to show that if $A$ is a $2 times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.










share|cite|improve this question















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Dec 9 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • Yes, it was a mistake
    – DIEGO R.
    Nov 26 at 15:03










  • You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
    – Sam Streeter
    Nov 26 at 15:12














4












4








4








This question already has an answer here:




  • All nilpotent $2times 2$ matrices

    2 answers




I have problems to show that if $A$ is a $2 times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.










share|cite|improve this question
















This question already has an answer here:




  • All nilpotent $2times 2$ matrices

    2 answers




I have problems to show that if $A$ is a $2 times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.





This question already has an answer here:




  • All nilpotent $2times 2$ matrices

    2 answers








linear-algebra matrices nilpotence






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edited Nov 26 at 18:55









Martin Sleziak

44.6k7115270




44.6k7115270










asked Nov 26 at 15:00









DIEGO R.

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Dec 9 at 16:24


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Dec 9 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Yes, it was a mistake
    – DIEGO R.
    Nov 26 at 15:03










  • You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
    – Sam Streeter
    Nov 26 at 15:12


















  • Yes, it was a mistake
    – DIEGO R.
    Nov 26 at 15:03










  • You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
    – Sam Streeter
    Nov 26 at 15:12
















Yes, it was a mistake
– DIEGO R.
Nov 26 at 15:03




Yes, it was a mistake
– DIEGO R.
Nov 26 at 15:03












You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
– Sam Streeter
Nov 26 at 15:12




You will find your answer from the answer(s) to this question: math.stackexchange.com/questions/1200829/…
– Sam Streeter
Nov 26 at 15:12










9 Answers
9






active

oldest

votes


















2














The only eigenvalue of $A$ is $0$: if $Av=lambda v$, then $A^nv=lambda^nv$.



Therefore
$$
a_{11}+a_{22}=0
$$

Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.



Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.



Now compute $A^2$.






share|cite|improve this answer





























    1














    If $A$ is a nilpotent $2 times 2$- matrix, then $0$ is the only eigenvalue of $A$.



    ($Ax = mu x$ implies $0=A^2x= mu^2x$).



    Hence the char. Polynomial ist $p(mu)=mu^2$. Now invoke Cayley-Hamilton.



    A direct proof is also possible: let $A=begin{pmatrix}a&b\c&dend{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.






    share|cite|improve this answer

















    • 1




      Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
      – DIEGO R.
      Nov 26 at 15:38





















    1














    If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $kleqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =det(A-xI)$, $deg p = 2$. Since $m mid p$, $deg m leqslant 2$, i.e. $k leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.






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      0














      Let A be $2 times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.






      share|cite|improve this answer





























        0














        Let $A$ be any matrix of any dimension $n$. Then
        $$
        mathbb{R}^n supseteq text{Im}(A) supseteq text{Im}(A^2) supseteq text{Im}(A^3) supseteq ldots
        $$



        If any containment in this chain is ever an equality, then the rest are also equalities.



        If $A$ is nilpotent, then this chain reaches zero when we get to $text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m times m$ for $m < n$ it must be zero by step $m$.






        share|cite|improve this answer





























          0














          Let $(lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then



          $$ 0 = 0v = A^nv = lambda^nv.$$



          This implies that $lambda = 0$. Since $lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are



          $$m_A(x) = xqquad m_A(x) = x^2$$



          and so $A = 0$ or $A^2 = 0$.






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            0














            $A^2-Tr(A)A+det(A)I=0 Rightarrow$



            $ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0cdot A^{n-2=}=0Rightarrow $



            $-Tr(A)A^{n-2}=0Rightarrow A^{n-2}=0.... Rightarrow A^2=0$



            (if $Tr(A)=0$ then from the initial relation $A^2=0$)






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              0














              If the eigenvalues of $A$ are $lambda$s, the eigenvalues of $A^n$ are $lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$lambda^n=0to lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$






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                0














                There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:



                We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.



                Since we're dealing with 2-dimensional space, there must be some $v_1$ such that ${v_0,v_1}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.



                This proof can be adapted to show that for any matrix of size $k times k$, if $A^n=0$, then $A^k=0$.






                share|cite|improve this answer




























                  9 Answers
                  9






                  active

                  oldest

                  votes








                  9 Answers
                  9






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  2














                  The only eigenvalue of $A$ is $0$: if $Av=lambda v$, then $A^nv=lambda^nv$.



                  Therefore
                  $$
                  a_{11}+a_{22}=0
                  $$

                  Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.



                  Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.



                  Now compute $A^2$.






                  share|cite|improve this answer


























                    2














                    The only eigenvalue of $A$ is $0$: if $Av=lambda v$, then $A^nv=lambda^nv$.



                    Therefore
                    $$
                    a_{11}+a_{22}=0
                    $$

                    Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.



                    Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.



                    Now compute $A^2$.






                    share|cite|improve this answer
























                      2












                      2








                      2






                      The only eigenvalue of $A$ is $0$: if $Av=lambda v$, then $A^nv=lambda^nv$.



                      Therefore
                      $$
                      a_{11}+a_{22}=0
                      $$

                      Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.



                      Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.



                      Now compute $A^2$.






                      share|cite|improve this answer












                      The only eigenvalue of $A$ is $0$: if $Av=lambda v$, then $A^nv=lambda^nv$.



                      Therefore
                      $$
                      a_{11}+a_{22}=0
                      $$

                      Also the determinant is $0$, so $a_{11}a_{22}-a_{12}a_{21}=0$.



                      Hence $a_{22}=-a_{11}$ and $a_{12}a_{21}=-a_{11}^2$.



                      Now compute $A^2$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 26 at 15:57









                      egreg

                      177k1484200




                      177k1484200























                          1














                          If $A$ is a nilpotent $2 times 2$- matrix, then $0$ is the only eigenvalue of $A$.



                          ($Ax = mu x$ implies $0=A^2x= mu^2x$).



                          Hence the char. Polynomial ist $p(mu)=mu^2$. Now invoke Cayley-Hamilton.



                          A direct proof is also possible: let $A=begin{pmatrix}a&b\c&dend{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.






                          share|cite|improve this answer

















                          • 1




                            Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                            – DIEGO R.
                            Nov 26 at 15:38


















                          1














                          If $A$ is a nilpotent $2 times 2$- matrix, then $0$ is the only eigenvalue of $A$.



                          ($Ax = mu x$ implies $0=A^2x= mu^2x$).



                          Hence the char. Polynomial ist $p(mu)=mu^2$. Now invoke Cayley-Hamilton.



                          A direct proof is also possible: let $A=begin{pmatrix}a&b\c&dend{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.






                          share|cite|improve this answer

















                          • 1




                            Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                            – DIEGO R.
                            Nov 26 at 15:38
















                          1












                          1








                          1






                          If $A$ is a nilpotent $2 times 2$- matrix, then $0$ is the only eigenvalue of $A$.



                          ($Ax = mu x$ implies $0=A^2x= mu^2x$).



                          Hence the char. Polynomial ist $p(mu)=mu^2$. Now invoke Cayley-Hamilton.



                          A direct proof is also possible: let $A=begin{pmatrix}a&b\c&dend{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.






                          share|cite|improve this answer












                          If $A$ is a nilpotent $2 times 2$- matrix, then $0$ is the only eigenvalue of $A$.



                          ($Ax = mu x$ implies $0=A^2x= mu^2x$).



                          Hence the char. Polynomial ist $p(mu)=mu^2$. Now invoke Cayley-Hamilton.



                          A direct proof is also possible: let $A=begin{pmatrix}a&b\c&dend{pmatrix}$, compute $A^2$ an determine $a,b,c$ and $d$ via $A^2=0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 26 at 15:13









                          Fred

                          44.2k1845




                          44.2k1845








                          • 1




                            Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                            – DIEGO R.
                            Nov 26 at 15:38
















                          • 1




                            Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                            – DIEGO R.
                            Nov 26 at 15:38










                          1




                          1




                          Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                          – DIEGO R.
                          Nov 26 at 15:38






                          Dear Fred. The hypotesis is $A^{n}$ iquals zero, not $A^{2}$. But thanks, you gave me the idea of use the unique proper value and the cayley hamilton theorem.
                          – DIEGO R.
                          Nov 26 at 15:38













                          1














                          If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $kleqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =det(A-xI)$, $deg p = 2$. Since $m mid p$, $deg m leqslant 2$, i.e. $k leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.






                          share|cite|improve this answer


























                            1














                            If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $kleqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =det(A-xI)$, $deg p = 2$. Since $m mid p$, $deg m leqslant 2$, i.e. $k leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.






                            share|cite|improve this answer
























                              1












                              1








                              1






                              If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $kleqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =det(A-xI)$, $deg p = 2$. Since $m mid p$, $deg m leqslant 2$, i.e. $k leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.






                              share|cite|improve this answer












                              If $A^n =O$, then $f(x) = x^n$ is an annihilating polynomial of $A$. Hence the minimal polynomial of $A$ is $m(x)= x^k$ for some $kleqslant n$. By Hamilton-Cayley theorem, the characteristic polynomial $p(x) $ of $A$ annihilates $A$. Since $p(x) =det(A-xI)$, $deg p = 2$. Since $m mid p$, $deg m leqslant 2$, i.e. $k leqslant 2$. Then either $m(x) =x$ or $m(x)=x^2$. Therefore $m(A) = A = O$ or $m(A) = A^2 = O$. Either case, $A^2 =O$.







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                              answered Nov 26 at 15:18









                              xbh

                              5,6551522




                              5,6551522























                                  0














                                  Let A be $2 times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.






                                  share|cite|improve this answer


























                                    0














                                    Let A be $2 times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      Let A be $2 times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.






                                      share|cite|improve this answer












                                      Let A be $2 times 2$ matrix such that $A^n$=0, for some n. Then clearly the matrix A satisfies the polynomial p(x)=$x^n$. So characteristic polynomial of A will annihilate p(x ) so using division algorithm it will be $x^2$ (since a degree characteristic polynomial is equal to the order of matrix ). So by Cayley Hamilton theorem, matrix A should satisfy the characteristic polynomial so $A^2=0$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 26 at 15:54









                                      MANI SHANKAR PANDEY

                                      276




                                      276























                                          0














                                          Let $A$ be any matrix of any dimension $n$. Then
                                          $$
                                          mathbb{R}^n supseteq text{Im}(A) supseteq text{Im}(A^2) supseteq text{Im}(A^3) supseteq ldots
                                          $$



                                          If any containment in this chain is ever an equality, then the rest are also equalities.



                                          If $A$ is nilpotent, then this chain reaches zero when we get to $text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m times m$ for $m < n$ it must be zero by step $m$.






                                          share|cite|improve this answer


























                                            0














                                            Let $A$ be any matrix of any dimension $n$. Then
                                            $$
                                            mathbb{R}^n supseteq text{Im}(A) supseteq text{Im}(A^2) supseteq text{Im}(A^3) supseteq ldots
                                            $$



                                            If any containment in this chain is ever an equality, then the rest are also equalities.



                                            If $A$ is nilpotent, then this chain reaches zero when we get to $text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m times m$ for $m < n$ it must be zero by step $m$.






                                            share|cite|improve this answer
























                                              0












                                              0








                                              0






                                              Let $A$ be any matrix of any dimension $n$. Then
                                              $$
                                              mathbb{R}^n supseteq text{Im}(A) supseteq text{Im}(A^2) supseteq text{Im}(A^3) supseteq ldots
                                              $$



                                              If any containment in this chain is ever an equality, then the rest are also equalities.



                                              If $A$ is nilpotent, then this chain reaches zero when we get to $text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m times m$ for $m < n$ it must be zero by step $m$.






                                              share|cite|improve this answer












                                              Let $A$ be any matrix of any dimension $n$. Then
                                              $$
                                              mathbb{R}^n supseteq text{Im}(A) supseteq text{Im}(A^2) supseteq text{Im}(A^3) supseteq ldots
                                              $$



                                              If any containment in this chain is ever an equality, then the rest are also equalities.



                                              If $A$ is nilpotent, then this chain reaches zero when we get to $text{Im}(A^n)$. By the above paragraph, it must strictly decrease at each step. Consequently, if $A$ is $m times m$ for $m < n$ it must be zero by step $m$.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Nov 26 at 15:59









                                              hunter

                                              14k22437




                                              14k22437























                                                  0














                                                  Let $(lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then



                                                  $$ 0 = 0v = A^nv = lambda^nv.$$



                                                  This implies that $lambda = 0$. Since $lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are



                                                  $$m_A(x) = xqquad m_A(x) = x^2$$



                                                  and so $A = 0$ or $A^2 = 0$.






                                                  share|cite|improve this answer


























                                                    0














                                                    Let $(lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then



                                                    $$ 0 = 0v = A^nv = lambda^nv.$$



                                                    This implies that $lambda = 0$. Since $lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are



                                                    $$m_A(x) = xqquad m_A(x) = x^2$$



                                                    and so $A = 0$ or $A^2 = 0$.






                                                    share|cite|improve this answer
























                                                      0












                                                      0








                                                      0






                                                      Let $(lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then



                                                      $$ 0 = 0v = A^nv = lambda^nv.$$



                                                      This implies that $lambda = 0$. Since $lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are



                                                      $$m_A(x) = xqquad m_A(x) = x^2$$



                                                      and so $A = 0$ or $A^2 = 0$.






                                                      share|cite|improve this answer












                                                      Let $(lambda, v)$ be an eigenvalue-eigenvector pair for $A$. Then



                                                      $$ 0 = 0v = A^nv = lambda^nv.$$



                                                      This implies that $lambda = 0$. Since $lambda$ is an arbitrary eigenvalue, the characteristic polynomial of $A$ is $p_A(x) = x^2$. Thus, the possibilities for the minimal polynomial are



                                                      $$m_A(x) = xqquad m_A(x) = x^2$$



                                                      and so $A = 0$ or $A^2 = 0$.







                                                      share|cite|improve this answer












                                                      share|cite|improve this answer



                                                      share|cite|improve this answer










                                                      answered Nov 26 at 16:18









                                                      Santana Afton

                                                      2,5592629




                                                      2,5592629























                                                          0














                                                          $A^2-Tr(A)A+det(A)I=0 Rightarrow$



                                                          $ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0cdot A^{n-2=}=0Rightarrow $



                                                          $-Tr(A)A^{n-2}=0Rightarrow A^{n-2}=0.... Rightarrow A^2=0$



                                                          (if $Tr(A)=0$ then from the initial relation $A^2=0$)






                                                          share|cite|improve this answer


























                                                            0














                                                            $A^2-Tr(A)A+det(A)I=0 Rightarrow$



                                                            $ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0cdot A^{n-2=}=0Rightarrow $



                                                            $-Tr(A)A^{n-2}=0Rightarrow A^{n-2}=0.... Rightarrow A^2=0$



                                                            (if $Tr(A)=0$ then from the initial relation $A^2=0$)






                                                            share|cite|improve this answer
























                                                              0












                                                              0








                                                              0






                                                              $A^2-Tr(A)A+det(A)I=0 Rightarrow$



                                                              $ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0cdot A^{n-2=}=0Rightarrow $



                                                              $-Tr(A)A^{n-2}=0Rightarrow A^{n-2}=0.... Rightarrow A^2=0$



                                                              (if $Tr(A)=0$ then from the initial relation $A^2=0$)






                                                              share|cite|improve this answer












                                                              $A^2-Tr(A)A+det(A)I=0 Rightarrow$



                                                              $ A^n-Tr(A)A^{n-1}+det(A)A^{n-2}=0cdot A^{n-2=}=0Rightarrow $



                                                              $-Tr(A)A^{n-2}=0Rightarrow A^{n-2}=0.... Rightarrow A^2=0$



                                                              (if $Tr(A)=0$ then from the initial relation $A^2=0$)







                                                              share|cite|improve this answer












                                                              share|cite|improve this answer



                                                              share|cite|improve this answer










                                                              answered Nov 26 at 16:24









                                                              giannispapav

                                                              1,534324




                                                              1,534324























                                                                  0














                                                                  If the eigenvalues of $A$ are $lambda$s, the eigenvalues of $A^n$ are $lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$lambda^n=0to lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$






                                                                  share|cite|improve this answer


























                                                                    0














                                                                    If the eigenvalues of $A$ are $lambda$s, the eigenvalues of $A^n$ are $lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$lambda^n=0to lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$






                                                                    share|cite|improve this answer
























                                                                      0












                                                                      0








                                                                      0






                                                                      If the eigenvalues of $A$ are $lambda$s, the eigenvalues of $A^n$ are $lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$lambda^n=0to lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$






                                                                      share|cite|improve this answer












                                                                      If the eigenvalues of $A$ are $lambda$s, the eigenvalues of $A^n$ are $lambda^n$s therefore according to Cayley-Hamilton's theorem we obtain $$lambda^n=0to lambda=0$$therefore all the eigenvalues of $A$ are zero and the characteristic would become$$lambda^2=0$$again by applying Cayley-Hamilton's theorem we finally have$$A^2=0$$







                                                                      share|cite|improve this answer












                                                                      share|cite|improve this answer



                                                                      share|cite|improve this answer










                                                                      answered Nov 26 at 17:20









                                                                      Mostafa Ayaz

                                                                      13.7k3836




                                                                      13.7k3836























                                                                          0














                                                                          There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:



                                                                          We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.



                                                                          Since we're dealing with 2-dimensional space, there must be some $v_1$ such that ${v_0,v_1}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.



                                                                          This proof can be adapted to show that for any matrix of size $k times k$, if $A^n=0$, then $A^k=0$.






                                                                          share|cite|improve this answer


























                                                                            0














                                                                            There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:



                                                                            We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.



                                                                            Since we're dealing with 2-dimensional space, there must be some $v_1$ such that ${v_0,v_1}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.



                                                                            This proof can be adapted to show that for any matrix of size $k times k$, if $A^n=0$, then $A^k=0$.






                                                                            share|cite|improve this answer
























                                                                              0












                                                                              0








                                                                              0






                                                                              There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:



                                                                              We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.



                                                                              Since we're dealing with 2-dimensional space, there must be some $v_1$ such that ${v_0,v_1}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.



                                                                              This proof can be adapted to show that for any matrix of size $k times k$, if $A^n=0$, then $A^k=0$.






                                                                              share|cite|improve this answer












                                                                              There must be some vector $v_0$ such that $Av_0=0$. If that's not intuitively obvious, here's a proof:



                                                                              We have that $A^n=0$ for all $v$. We can rewrite that as $A(A^{n-1}v)=0$ for all $v$. Thus, we can pick any vector to be $v$, and defining $v_0=A^{n-1}v$ will give us a vector such that $Av_0=0$.



                                                                              Since we're dealing with 2-dimensional space, there must be some $v_1$ such that ${v_0,v_1}$ is a basis for the space. So we can write $Av_1$ in terms of those vectors: $Av_1=c_0v_0+c_1v_1$ for some $c_0,c_1$. If we define $v_2=Av_1$, by the linearity of matrix multiplication, we have we have that $Av_2=A(c_0v_0+c_1v_0)=c_0Av_0+c_1Av_1=0+c_1(c_0v_0+c_1v_1)=c_1Av_1$. So there is some real number such that $Av_2=c_1v_2$ (that is, $v_2$ is an eigenvector with eigenvalue $c_1$). So now $A^nv_1=A^{n-1}(Av_1)=A^{n-1}v_2=c_1^{n-1}v_2$. Since we have to have $A^b=0$, we have to have $c_1^{n-1}v_2=0$. And if $c_1^{n-1}v_2=0$, then it follows that $c_1v_2=0$. So $A^2$ sends both basis vectors to the zero vector, hence it is the zero matrix.



                                                                              This proof can be adapted to show that for any matrix of size $k times k$, if $A^n=0$, then $A^k=0$.







                                                                              share|cite|improve this answer












                                                                              share|cite|improve this answer



                                                                              share|cite|improve this answer










                                                                              answered Nov 26 at 18:11









                                                                              Acccumulation

                                                                              6,7862617




                                                                              6,7862617















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