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Let $A \in \mathbb{R}^{n\times m}$. Define

$$\|A\| = \sum_{i=1}^{\min(n,m)} \sigma_i$$

where $\sigma_i$ are the singular values of $A$. How to show that $\|A\|$ is norm?

We want to verify the definition of norm.

Homogeneity $\|\alpha A\| = |\alpha| \times \|A\|$ for $\alpha \in \mathbb{R}$

Triangular Inequality: $\|A+B\| \leq \|A\| + \|B\|$

Positive definiteness: if $\|A\|=0$, then $A=0$

Rodrigo de Azevedo
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Josephin
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    Are you of your formula? Normally, $;|A| = \sqrt{\sum_{i=1}^{\min(n,m)} \sigma_i^2}$. – Bernard Nov 04 '20 at 10:04
  • The main (only) difficulty concerns the triangle inequality. It was already demonstrated, for example here: https://math.stackexchange.com/q/564873/621834 – Damien Nov 04 '20 at 10:14
  • @Jospehin The norm $\sum_i \sigma_i$ is not called the "Frobenius norm", so it is not clear what you are asking. Please clarify whether you are asking about the Frobenius norm or about the norm with the formula that you give (usually called the "nuclear norm"). – Ben Grossmann Nov 04 '20 at 13:20
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    Homogeneity and positive definiteness are obvious. To prove the triangle inequality, since the appendment of zero rows or columns does not affect the nuclear norm, you may assume that $A$ is a square matrix. Then you may use the characterisation $|A|=\max|\operatorname{tr}(AU)|$ where the maximum is taken over all unitary matrices $U$. – user1551 Nov 07 '20 at 09:06

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