How to find dimension of eigenspace?












0












$begingroup$


Given $lambda$ = 2 and matrix A:



A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.



$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$



"This implies that $x_2$ = 0".



How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$

computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 20:22










  • $begingroup$
    The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
    $endgroup$
    – Bernard
    Dec 4 '18 at 20:27










  • $begingroup$
    @Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
    $endgroup$
    – Federico
    Dec 4 '18 at 20:27










  • $begingroup$
    $dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
    $endgroup$
    – Federico
    Dec 4 '18 at 20:29










  • $begingroup$
    @Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 21:25


















0












$begingroup$


Given $lambda$ = 2 and matrix A:



A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.



$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$



"This implies that $x_2$ = 0".



How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$

computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.






linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 20:22










  • $begingroup$
    The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
    $endgroup$
    – Bernard
    Dec 4 '18 at 20:27










  • $begingroup$
    @Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
    $endgroup$
    – Federico
    Dec 4 '18 at 20:27










  • $begingroup$
    $dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
    $endgroup$
    – Federico
    Dec 4 '18 at 20:29










  • $begingroup$
    @Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 21:25
















0












0








0





$begingroup$


Given $lambda$ = 2 and matrix A:



A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.



$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$



"This implies that $x_2$ = 0".



How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$

computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.






linear-algebra







share|cite|improve this question









$endgroup$




Given $lambda$ = 2 and matrix A:



A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$



My textbook says that to find the eigenvectors of $lambda = 2$, solve the homogeneous linear system represented by $(2I - A)x = 0$.



$2I - A$ =
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$



"This implies that $x_2$ = 0".



How is the matrix
$begin{bmatrix}
0 & -1 & 0 \
0 & 0 & 0 \
0 & 0 & 0 \
end{bmatrix}$

computed and how does this imply that $x_2$ = 0? My textbook does not give an explanation.






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 20:18









Evan KimEvan Kim

998




998












  • $begingroup$
    The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 20:22










  • $begingroup$
    The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
    $endgroup$
    – Bernard
    Dec 4 '18 at 20:27










  • $begingroup$
    @Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
    $endgroup$
    – Federico
    Dec 4 '18 at 20:27










  • $begingroup$
    $dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
    $endgroup$
    – Federico
    Dec 4 '18 at 20:29










  • $begingroup$
    @Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 21:25




















  • $begingroup$
    The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
    $endgroup$
    – Michael Burr
    Dec 4 '18 at 20:22










  • $begingroup$
    The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
    $endgroup$
    – Bernard
    Dec 4 '18 at 20:27










  • $begingroup$
    @Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
    $endgroup$
    – Federico
    Dec 4 '18 at 20:27










  • $begingroup$
    $dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
    $endgroup$
    – Federico
    Dec 4 '18 at 20:29










  • $begingroup$
    @Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
    $endgroup$
    – Evan Kim
    Dec 4 '18 at 21:25


















$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22




$begingroup$
The matrix that you're asking about is really $2I-A$, just the operations directly on the matrices, there's not really anything to the computation...
$endgroup$
– Michael Burr
Dec 4 '18 at 20:22












$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27




$begingroup$
The rank of $2I-A$ is the codimension of the eigenspace $E_2$.
$endgroup$
– Bernard
Dec 4 '18 at 20:27












$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27




$begingroup$
@Evan Kim This is the second question regarding this matrix and this exercise in general. Maybe you should read again about eigenvalues, characteristic polynomial, kernel, rank, dimension... It seems that the definitions are not completely clear. I feel it would be better for you if you first go back to study the concepts
$endgroup$
– Federico
Dec 4 '18 at 20:27












$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29




$begingroup$
$dim ker(2I-A) = 3-mathrm{rank}(2I-A)=3-1=2$
$endgroup$
– Federico
Dec 4 '18 at 20:29












$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25






$begingroup$
@Federico The issue is that I am having a difficult time grasping the definitions in the study material assigned to me in class. I do agree that these are trivial questions that should be self-explanatory though yet I have still struggled the entire semester. An example is the book explains rank and dimension and I understand that, but it fails to provide a connection between concepts in different chapters and perhaps expect us to see these connections by ourselves.
$endgroup$
– Evan Kim
Dec 4 '18 at 21:25












3 Answers
3






active

oldest

votes


















2












$begingroup$

The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    We have that



    $$begin{bmatrix}
    0 & -1 & 0 \
    0 & 0 & 0 \
    0 & 0 & 0 \
    end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$



    therefore what is the dimension of the eigenspace?






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      You must find all a basis for the space of vectors such that:



      $Amathbf v = lambda mathbf v$



      and with a little algebra
      $(A- lambda I)mathbf v = mathbf 0$



      If $x_2 ne 0$



      $(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
        If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
          If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
            If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?






            share|cite|improve this answer









            $endgroup$



            The matrix $2I-A$ is simply computed as $$begin{bmatrix}2 & 0 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}-begin{bmatrix}2 & 1 & 0\0 & 2 & 0\0 & 0 & 2end{bmatrix}=begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}$$
            If a vector $x$ with coordinates $x_1,x_2,x_3$ satisfies $(2I-A)x=0$, then it must be that $x_2=0$. This is because $$begin{bmatrix}0 & -1 & 0\0 & 0 & 0\0 & 0 & 0end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}implies begin{bmatrix}-x_2\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$$ so we must have $x_2=0$. Notice that $x_1$ and $x_3$ can be any number, so what does this tell you about the dimension of the eigenspace (i.e. the dimension of ${xinmathbb C^3:(2A-I)x=0}$)?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 4 '18 at 20:31









            DaveDave

            8,76711033




            8,76711033























                2












                $begingroup$

                We have that



                $$begin{bmatrix}
                0 & -1 & 0 \
                0 & 0 & 0 \
                0 & 0 & 0 \
                end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$



                therefore what is the dimension of the eigenspace?






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  We have that



                  $$begin{bmatrix}
                  0 & -1 & 0 \
                  0 & 0 & 0 \
                  0 & 0 & 0 \
                  end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$



                  therefore what is the dimension of the eigenspace?






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    We have that



                    $$begin{bmatrix}
                    0 & -1 & 0 \
                    0 & 0 & 0 \
                    0 & 0 & 0 \
                    end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$



                    therefore what is the dimension of the eigenspace?






                    share|cite|improve this answer









                    $endgroup$



                    We have that



                    $$begin{bmatrix}
                    0 & -1 & 0 \
                    0 & 0 & 0 \
                    0 & 0 & 0 \
                    end{bmatrix}vec x=vec 0 implies vec x=(x_1,0,x_3)$$



                    therefore what is the dimension of the eigenspace?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 4 '18 at 20:20









                    gimusigimusi

                    92.9k94494




                    92.9k94494























                        0












                        $begingroup$

                        You must find all a basis for the space of vectors such that:



                        $Amathbf v = lambda mathbf v$



                        and with a little algebra
                        $(A- lambda I)mathbf v = mathbf 0$



                        If $x_2 ne 0$



                        $(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          You must find all a basis for the space of vectors such that:



                          $Amathbf v = lambda mathbf v$



                          and with a little algebra
                          $(A- lambda I)mathbf v = mathbf 0$



                          If $x_2 ne 0$



                          $(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            You must find all a basis for the space of vectors such that:



                            $Amathbf v = lambda mathbf v$



                            and with a little algebra
                            $(A- lambda I)mathbf v = mathbf 0$



                            If $x_2 ne 0$



                            $(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$






                            share|cite|improve this answer









                            $endgroup$



                            You must find all a basis for the space of vectors such that:



                            $Amathbf v = lambda mathbf v$



                            and with a little algebra
                            $(A- lambda I)mathbf v = mathbf 0$



                            If $x_2 ne 0$



                            $(A- 2I)mathbf v = begin{bmatrix} x_2\0\0 end{bmatrix} ne mathbf 0$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 4 '18 at 20:39









                            Doug MDoug M

                            44.5k31854




                            44.5k31854






























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                                Are they similar matrix

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                                0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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