# Question about singular values, traces and determinants of a matrix

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.

• first of all,why are there absolute values of singular values?? Jun 20, 2022 at 19:18
• Claim (2) is not true without further assumptions on $A$ (a counterexample is $A=\begin{bmatrix} 1 \\ & -1 \end{bmatrix}$). Jun 20, 2022 at 19:18
• Is there something missing from the problem statement? Why is the statement about orthonormality included if the question doesn't mention orthonormality elsewhere? Jun 20, 2022 at 19:20
• 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 Jun 20, 2022 at 19:21
• (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 $A \in \mathbb C^{n\times n}$. Jun 20, 2022 at 22:43

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.