Distinct eigenvalues implies $A in mathbb{R}^{n times n}$ is diagonalisable












2












$begingroup$


Theorem:




If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.




Proof:



Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.



Any hint(s) to assist me would be helpful.



Thanks in advance.






linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Theorem:




    If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.




    Proof:



    Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
    Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.



    Any hint(s) to assist me would be helpful.



    Thanks in advance.






    linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Theorem:




      If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.




      Proof:



      Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
      Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.



      Any hint(s) to assist me would be helpful.



      Thanks in advance.






      linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions







      share|cite|improve this question











      $endgroup$




      Theorem:




      If an $n times n$ matrix has n distinct eigenvalues then A is diagonalisable.




      Proof:



      Let $A in mathbb{R}^{n times n}$. Suppose A is not diagonalisable.
      Then, by definition, for a given $D in mathbb{R}^{n times n}$ there exists no invertible matrix $P in mathbb{R}^{n times n}$ such that $P^{-1}AP = D$.



      Any hint(s) to assist me would be helpful.



      Thanks in advance.






      linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions




      linear-algebra proof-verification eigenvalues-eigenvectors eigenfunctions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 16 '18 at 2:55









      JimmyK4542

      41.1k245107




      41.1k245107










      asked Dec 16 '18 at 2:52









      MathematicingMathematicing

      2,45521856




      2,45521856






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          Hint: I would prefer to prove it directly.



          Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.




          Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            don't you need symmetry for linear independent eigenvectors?
            $endgroup$
            – LinAlg
            Dec 16 '18 at 3:09






          • 1




            $begingroup$
            No. See this: math.stackexchange.com/a/29374
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:20










          • $begingroup$
            @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:28






          • 1




            $begingroup$
            Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:32










          • $begingroup$
            @ChrisCuster Satisfied.
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:35











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          Hint: I would prefer to prove it directly.



          Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.




          Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            don't you need symmetry for linear independent eigenvectors?
            $endgroup$
            – LinAlg
            Dec 16 '18 at 3:09






          • 1




            $begingroup$
            No. See this: math.stackexchange.com/a/29374
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:20










          • $begingroup$
            @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:28






          • 1




            $begingroup$
            Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:32










          • $begingroup$
            @ChrisCuster Satisfied.
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:35
















          4












          $begingroup$

          Hint: I would prefer to prove it directly.



          Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.




          Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            don't you need symmetry for linear independent eigenvectors?
            $endgroup$
            – LinAlg
            Dec 16 '18 at 3:09






          • 1




            $begingroup$
            No. See this: math.stackexchange.com/a/29374
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:20










          • $begingroup$
            @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:28






          • 1




            $begingroup$
            Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:32










          • $begingroup$
            @ChrisCuster Satisfied.
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:35














          4












          4








          4





          $begingroup$

          Hint: I would prefer to prove it directly.



          Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.




          Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.







          share|cite|improve this answer









          $endgroup$



          Hint: I would prefer to prove it directly.



          Eigenvectors for distinct eigenvalues are linearly independent. Thus we have a basis for $V$ consisting of eigenvectors. Simply let $P$ be the matrix whose columns are the basis vectors.




          Then $P^{-1}AP=D$, where $D$ is diagonal, and the entries on the diagonal are the eigenvalues.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 3:04









          Chris CusterChris Custer

          13.7k3827




          13.7k3827












          • $begingroup$
            don't you need symmetry for linear independent eigenvectors?
            $endgroup$
            – LinAlg
            Dec 16 '18 at 3:09






          • 1




            $begingroup$
            No. See this: math.stackexchange.com/a/29374
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:20










          • $begingroup$
            @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:28






          • 1




            $begingroup$
            Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:32










          • $begingroup$
            @ChrisCuster Satisfied.
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:35


















          • $begingroup$
            don't you need symmetry for linear independent eigenvectors?
            $endgroup$
            – LinAlg
            Dec 16 '18 at 3:09






          • 1




            $begingroup$
            No. See this: math.stackexchange.com/a/29374
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:20










          • $begingroup$
            @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:28






          • 1




            $begingroup$
            Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
            $endgroup$
            – Chris Custer
            Dec 16 '18 at 3:32










          • $begingroup$
            @ChrisCuster Satisfied.
            $endgroup$
            – Mathematicing
            Dec 16 '18 at 3:35
















          $begingroup$
          don't you need symmetry for linear independent eigenvectors?
          $endgroup$
          – LinAlg
          Dec 16 '18 at 3:09




          $begingroup$
          don't you need symmetry for linear independent eigenvectors?
          $endgroup$
          – LinAlg
          Dec 16 '18 at 3:09




          1




          1




          $begingroup$
          No. See this: math.stackexchange.com/a/29374
          $endgroup$
          – Chris Custer
          Dec 16 '18 at 3:20




          $begingroup$
          No. See this: math.stackexchange.com/a/29374
          $endgroup$
          – Chris Custer
          Dec 16 '18 at 3:20












          $begingroup$
          @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
          $endgroup$
          – Mathematicing
          Dec 16 '18 at 3:28




          $begingroup$
          @ChrisCuster By construction, the j column of the matrix P is the j eigenvectors in the L.I set S of eigenvectors. Clearly, set S is the column space of P. I'd like to show that every column of P has a pivot position. Can you give me a hint on this?
          $endgroup$
          – Mathematicing
          Dec 16 '18 at 3:28




          1




          1




          $begingroup$
          Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
          $endgroup$
          – Chris Custer
          Dec 16 '18 at 3:32




          $begingroup$
          Hmm. Clearly the columns are independent, as they are the elements of a basis. To get pivots you would have to take the transpose and row-reduce, say.
          $endgroup$
          – Chris Custer
          Dec 16 '18 at 3:32












          $begingroup$
          @ChrisCuster Satisfied.
          $endgroup$
          – Mathematicing
          Dec 16 '18 at 3:35




          $begingroup$
          @ChrisCuster Satisfied.
          $endgroup$
          – Mathematicing
          Dec 16 '18 at 3:35


















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