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Question

Suppose $X\in \mathbb{R}^{M\times N}$

$\|X\|_*=\mathrm{trace}(\sqrt{X^*X})=\sum_i^{\min{M,N}}\sigma_i$ where $\sigma_i$ is the singular values of $X$.

I know that $\mathrm{trace}({X^*X})=\sum_i^{\min{M,N}}\sigma_i^2$. Thus, the following equation holds.

$\sqrt{\mathrm{trace}({X^*X})}=\sqrt{\sum_i^{\min{M,N}}\sigma_i^2}$.

But I wonder how to prove $\mathrm{trace}(\sqrt{X^*X})=\sum_i^{\min{M,N}}\sigma_i$

Simply use the spectral decomposition theorem, so that $X = U\Sigma V^*$. — user91011, Oct 09 '13 at 16:31

score 1 · Accepted Answer · answered Oct 09 '13 at 16:31

Let $X = U \Sigma V^*$, where $U$ and $V$ are unitary and $\Sigma$ is nonnegative real diagonal, be an SVD of $X$. Then

$$X^*X = V \Sigma^2 V^*,$$

so

$$\sqrt{X^*X} = V \sqrt{\Sigma^2} V^* = V \Sigma V^*,$$

Notice that this is a similarity relation, so

$$\operatorname{tr} \sqrt{X^*X} = \operatorname{tr} (V \Sigma V^*) = \operatorname{tr} \Sigma.$$

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