Trace of symmetric matrix equals sum eigenvalues












3














I need to show that if $mathbf{S}$ is symmetric, then it's trace sums to the sum of the eigenvalues. But I don't know how to show this. Can anybody give me a hint?



P.S. Shame on my google skills, buy I really can't find any pages on this specific issue. Not with the assumption that $mathbf{S}$ is symmetric, and no proofs.






eigenvalues-eigenvectors trace symmetric-matrices







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  • 2




    This is true of any square matrix.
    – amd
    Nov 29 at 20:16
















3














I need to show that if $mathbf{S}$ is symmetric, then it's trace sums to the sum of the eigenvalues. But I don't know how to show this. Can anybody give me a hint?



P.S. Shame on my google skills, buy I really can't find any pages on this specific issue. Not with the assumption that $mathbf{S}$ is symmetric, and no proofs.






eigenvalues-eigenvectors trace symmetric-matrices







share|cite|improve this question


















  • 2




    This is true of any square matrix.
    – amd
    Nov 29 at 20:16














3












3








3







I need to show that if $mathbf{S}$ is symmetric, then it's trace sums to the sum of the eigenvalues. But I don't know how to show this. Can anybody give me a hint?



P.S. Shame on my google skills, buy I really can't find any pages on this specific issue. Not with the assumption that $mathbf{S}$ is symmetric, and no proofs.






eigenvalues-eigenvectors trace symmetric-matrices







share|cite|improve this question













I need to show that if $mathbf{S}$ is symmetric, then it's trace sums to the sum of the eigenvalues. But I don't know how to show this. Can anybody give me a hint?



P.S. Shame on my google skills, buy I really can't find any pages on this specific issue. Not with the assumption that $mathbf{S}$ is symmetric, and no proofs.






eigenvalues-eigenvectors trace symmetric-matrices




eigenvalues-eigenvectors trace symmetric-matrices






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asked Nov 29 at 13:28









Casper Thalen

1678




1678








  • 2




    This is true of any square matrix.
    – amd
    Nov 29 at 20:16














  • 2




    This is true of any square matrix.
    – amd
    Nov 29 at 20:16








2




2




This is true of any square matrix.
– amd
Nov 29 at 20:16




This is true of any square matrix.
– amd
Nov 29 at 20:16










4 Answers
4






active

oldest

votes


















4














If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$






share|cite|improve this answer

















  • 1




    Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
    – Casper Thalen
    Nov 29 at 13:37



















3














The trace can be expressed as
$$
operatorname{Tr}A=sum_ke_k'Ae_k,
$$
where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
$$
operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
$$






share|cite|improve this answer





















  • This is a very interesting approach, i didn't consider, thank you for this contribution!
    – Casper Thalen
    Nov 29 at 13:45



















2














Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}






share|cite|improve this answer































    1














    As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
    $(-1)^{n-1}cdotoperatorname{Tr}(A)$
    and $A$ is diagonalizable, the claim follows by Vieta.






    share|cite|improve this answer























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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$






      share|cite|improve this answer

















      • 1




        Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
        – Casper Thalen
        Nov 29 at 13:37
















      4














      If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$






      share|cite|improve this answer

















      • 1




        Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
        – Casper Thalen
        Nov 29 at 13:37














      4












      4








      4






      If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$






      share|cite|improve this answer












      If $S$ is a symmetric matrix then $S$ has a spectral decomposition as $S=PDP'$ where $D$ is the diagonal matrix consisting the eigenvalues of $S$ and $P$ is orthogonal. Then $tr(S)=tr(PDP')=tr(DP'P)=tr(D)=sum text{eigen values of } S.$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 29 at 13:34









      John_Wick

      1,309111




      1,309111








      • 1




        Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
        – Casper Thalen
        Nov 29 at 13:37














      • 1




        Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
        – Casper Thalen
        Nov 29 at 13:37








      1




      1




      Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
      – Casper Thalen
      Nov 29 at 13:37




      Yes, I litterally just figured this out! I came across this multiple times, but the step $tr(PDP')=tr(DP'P)$ seemed like a strech, as I've learned that order of multiplication matters in matrix algebra. However, when you realize that $tr(AB) = tr(BA)$ this step is easily shown to be justified! Thank you for your fast reply!
      – Casper Thalen
      Nov 29 at 13:37











      3














      The trace can be expressed as
      $$
      operatorname{Tr}A=sum_ke_k'Ae_k,
      $$
      where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
      $$
      operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
      $$






      share|cite|improve this answer





















      • This is a very interesting approach, i didn't consider, thank you for this contribution!
        – Casper Thalen
        Nov 29 at 13:45
















      3














      The trace can be expressed as
      $$
      operatorname{Tr}A=sum_ke_k'Ae_k,
      $$
      where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
      $$
      operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
      $$






      share|cite|improve this answer





















      • This is a very interesting approach, i didn't consider, thank you for this contribution!
        – Casper Thalen
        Nov 29 at 13:45














      3












      3








      3






      The trace can be expressed as
      $$
      operatorname{Tr}A=sum_ke_k'Ae_k,
      $$
      where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
      $$
      operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
      $$






      share|cite|improve this answer












      The trace can be expressed as
      $$
      operatorname{Tr}A=sum_ke_k'Ae_k,
      $$
      where $e_k$ are any orthonormal vectors. Since we assume that the matrix $A$ is symmetric, its eigenvectors are orthogonal. Suppose that $v_k$ are eigenvectors of $A$ with their corresponding eigenvalues $lambda_k$. Then
      $$
      operatorname{Tr}A=sum_kv_k'Av_k=sum_kv_k'lambda_kv_k=sum_klambda_kv_k'v_k=sum_klambda_k.
      $$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 29 at 13:44









      Cm7F7Bb

      12.4k32142




      12.4k32142












      • This is a very interesting approach, i didn't consider, thank you for this contribution!
        – Casper Thalen
        Nov 29 at 13:45


















      • This is a very interesting approach, i didn't consider, thank you for this contribution!
        – Casper Thalen
        Nov 29 at 13:45
















      This is a very interesting approach, i didn't consider, thank you for this contribution!
      – Casper Thalen
      Nov 29 at 13:45




      This is a very interesting approach, i didn't consider, thank you for this contribution!
      – Casper Thalen
      Nov 29 at 13:45











      2














      Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}






      share|cite|improve this answer




























        2














        Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}






        share|cite|improve this answer


























          2












          2








          2






          Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}






          share|cite|improve this answer














          Every symmetric real matrix $M$ is diagonalizable (spectral theorem). So, $M$ is similar to a diagonal matrix $D$ and thereforebegin{align}operatorname{tr}M&=operatorname{tr}D\&=sumtext{ eigenvalues of }\&=sumtext{ eigenvalues of }M.end{align}







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 at 21:33









          Glorfindel

          3,41981730




          3,41981730










          answered Nov 29 at 13:33









          José Carlos Santos

          148k22117218




          148k22117218























              1














              As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
              $(-1)^{n-1}cdotoperatorname{Tr}(A)$
              and $A$ is diagonalizable, the claim follows by Vieta.






              share|cite|improve this answer




























                1














                As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
                $(-1)^{n-1}cdotoperatorname{Tr}(A)$
                and $A$ is diagonalizable, the claim follows by Vieta.






                share|cite|improve this answer


























                  1












                  1








                  1






                  As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
                  $(-1)^{n-1}cdotoperatorname{Tr}(A)$
                  and $A$ is diagonalizable, the claim follows by Vieta.






                  share|cite|improve this answer














                  As the coefficient of $lambda^{n-1}$ in characteristic polynomial of $A$ is
                  $(-1)^{n-1}cdotoperatorname{Tr}(A)$
                  and $A$ is diagonalizable, the claim follows by Vieta.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 at 20:05

























                  answered Nov 29 at 16:05









                  Michael Hoppe

                  10.8k31834




                  10.8k31834






























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