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I'm self-studying linear algebra and found this problem on a friend's lecture notes.

Let $A \in \mathbb R^{n\times n}$ with singular values $\sigma_1, \sigma_2, \dots \sigma_n$. Recalling that $B\in\mathbb R^{n\times n}$ is orthonormal iff $B^T B = B B^T = I$, prove that if $A\succeq0$, then

$$ (1) \quad tr (A) = \sum^{n}_{i=1}|\sigma_i|$$ $$ (2) \quad det (A) = \prod^{n}_{i=1}\sigma_i$$

I know that by expanding the characteristic polynomial of $A$ one can find by comparison that $tr(A) = \sum^{n}_{i=1}\lambda_i$ and $det(A) = \prod^{n}_{i=1}\lambda_i$, where $\lambda_i$ are the eigenvalues of $A$. I don't know how to make the singular values of $A$ appear from here. I'm also not sure of this approach because I'm not using the hint.

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  • $\begingroup$ first of all,why are there absolute values of singular values?? $\endgroup$
    – Exodd
    Jun 20, 2022 at 19:18
  • $\begingroup$ Claim (2) is not true without further assumptions on $A$ (a counterexample is $A=\begin{bmatrix} 1 \\ & -1 \end{bmatrix}$). $\endgroup$
    – angryavian
    Jun 20, 2022 at 19:18
  • $\begingroup$ Is there something missing from the problem statement? Why is the statement about orthonormality included if the question doesn't mention orthonormality elsewhere? $\endgroup$
    – angryavian
    Jun 20, 2022 at 19:20
  • $\begingroup$ and also (1) cannot be true either. You can keep the diagonal of a matrix fixed and at the same time rise the norm as much as you want by changing just one offdiagonal element $\endgroup$
    – Exodd
    Jun 20, 2022 at 19:21
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    $\begingroup$ (1) holds iff $A$ is normal with non-negative eigenvalues... I.e. you've written $A\succeq \mathbf 0$ but this is ambiguous (hence claim may not be true depending on choice of definition) for real matrices. You need to explicitly state that $A$ is symmetric, or better change the field and let $A \in \mathbb C^{n\times n}$. $\endgroup$ Jun 20, 2022 at 22:43

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For this question one needs the assumption that $A$ is symmetric positive semi-definite(PSD) which I believe is what is meant by $ A \succeq 0$. Then we can invoke the spectral theorem which tells us that $A = P \Lambda P^T$ where $P$ has orthonormal columns corresponding to the eigenvectors of $A$ and $\Lambda$ is a diagonal matrix with eigenvalues on the diagonal. Note that we used the fact that for a matrix that is PSD the decomposition given by the spectral theorem is equivalent to an SVD. That is one example of an SVD given by $ A = U \Sigma V^T $ is satisfied by setting $U = V = P$ and $\Sigma = \Lambda$. From this we prove the properties as follows,

$$ tr(A) = tr(P \Lambda P^T) = tr(P^TP \Lambda) = tr(\Lambda) = \sum_{i=1}^{n} \lambda_i = \sum_{I=1}^{n} \sigma_i $$

$$ det(A) = det(P\Lambda P^T) = det(P)det(\Lambda)det(P^T) = det(P)det(\Lambda)\frac{1}{det(P)} = det(\Lambda) = \Pi_{i=1}^{n}\lambda_i = \Pi_{i=1}^{n} \sigma_i $$

An additional comment would be the absolute value bars in property $1$ are unnecessary because the singular values are always non-negative. We are essentially using the definitions of trace and determinant you provided, but I have added the calculation verifying that the change of basis did not change the trace or determinant. That is you could skip most of the equalities and jump straight to $tr(A) = \sum_{i=1}^{n} \lambda_i$ and then say the SVD given by the eigendecomposition justifies the last equality.

To see a counterexample to both claims when dropping the PSD assumption choose the diagonal 2-by-2 matrix with 1 and -1 on the diagonal.

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