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If $A^*\in S^n$, is a symmetric matrix, then is true that $$\mbox{ Tr }(A^*A)=\sum_{i=1}^n (\lambda_i)^2$$

I know, bt the spectral theorem, that For any symmetric matrix, there are eigenvalues $\lambda_1,\ldots,\lambda_n$, with corresponding eigenvectors $v_1,\ldots,v_n$ which are orthonormal. We can then write $$\sum_{i=1}^n v_i^T\lambda_i v_i=V\Lambda V,$$

where $V$ is the matrix with $v_i$’s arranged as column vectors and $\Lambda$ is the diagonal matrix of eigenvalues. But, is always true that $\mbox{ Tr }(A^*A)=\sum_{i=1}^n v_i^T(\lambda_i A)v_i$, is very weird, 'cause is this should be the square of the term of matrix $A*$. Thanks

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    $\begingroup$ (1) The middle equation should be $V \Lambda V^\top = \sum_{i=1}^n \lambda_i v_i v_i^\top$ (note the transposes). (2) If the $v_i$ are orthonormal, then can't the right-hand side of the first equation be simplified by noting $v_i^\top (\lambda_i A) v_i = v_i^\top \lambda_i^2 v_i = \lambda_i^2$? $\endgroup$
    – angryavian
    Nov 9, 2021 at 22:56
  • $\begingroup$ Are you specifically asking whether $\mbox{ Tr }(A^*A)=\sum_{i=1}^n v_i^T(\lambda_i A)v_i$ holds when $\lambda_i$ are the eigenvalues of $A$ and $\{v_1,\dots,v_n\}$ is an associated orthonormal eigenbasis? $\endgroup$ Nov 9, 2021 at 22:56
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    $\begingroup$ Please do not delete a post immediately after receiving an answer. This is disrespectful to those who took the time to help you improve your question, those who took the time to answer your question, and future readers who might find this Q&A helpful. $\endgroup$
    – Xander Henderson
    Nov 10, 2021 at 0:31
  • $\begingroup$ You don't need the spectral theorem. The trace of $A^\ast A$ is the same as the trace of $A^2$ because $A^\ast=A$. The trace of $A^2$ is the sum of all eigenvalues of $A^2$. The eigenvalues of $A^2$ are the squares of the eigenvalues of $A$. Therefore... $\endgroup$
    – user1551
    Nov 10, 2021 at 7:53

1 Answer 1

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If $A$ is a complex symmetric matrix, then $A=V\Lambda V^*$ for some unitary $V$ and diagonal $\Lambda$. (Note that the asterisk $*$ denotes conjugate transpose.)

Then $A^* A = V \Lambda^* \Lambda V^*$ and using the cyclic property of the trace, we have $$\operatorname{Tr}(A^* A) = \operatorname{Tr}(V \Lambda^* \Lambda V^*) = \operatorname{Tr}(\Lambda^* \Lambda V^* V) = \operatorname{Tr}(\Lambda^* \Lambda) = \sum_{i=1}^n \overline{\lambda_i} \lambda_i = \sum_{i=1}^n |\lambda_i|^2.$$

When $A$ is real symmetric, this reduces to $\operatorname{Tr}(A^\top A) = \sum_{i=1}^n \lambda_i^2$, and in the comments I mentioned that $v_i^\top (\lambda_i A) v_i = \lambda_i^2$.

(Note that the assumption that $\operatorname{Tr}(A)=1$ is not needed.)


Regarding the "very weird" comment in your last sentence:

You are correct that $\operatorname{Tr}(A^\top A)$ is the sum of squares of the entries of $A$. The above work shows that for symmetric matrices, it is also the sum of squares of the eigenvalues.

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