Trace identity for symmetric matrix.

Question

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

Answer 1

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.)

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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.