# Clarification on the Wikipedia existence proof of the SVD of a matrix using the Spectral Theorem

I am trying to understand the existence proof given by Wikipedia here of the SVD of matrix using the spectral theorem for Hermitian matrices. Suppose we have a complex matrix $$M$$ of dimension $$m \times n$$. Let $$V$$ be the matrix whose $$i$$'th column is the $$i$$'th eigenvector of $$M^*M$$. Write $$V = \begin{bmatrix} V_1 & V_2 \end{bmatrix}$$, where $$V_1$$ consists of the eigenvectors of $$M^*M$$ corresponding to non-zero eigenvalues, and $$V_2$$ the eigenvectors of corresponding to zero eigenvalues. The author then writes that $$V_2^*M^*MV_2 = 0$$ implies that $$MV_2 = 0$$.

The rational given is that $$trace(V_2^*M^*MV_2) = ||MV_2||^2$$, and $$||AA^t|| = 0 \iff A = 0$$ (trace norm), then the result follows. I don't understand the first statement here. Why is $$trace(V_2^*M^*MV_2) = ||MV_2||^2$$ ? I might be confusing the notation being used.

There is no "why". It's a definition. The norm is called Frobenius norm, not trace norm (the name "trace norm" refers to another matrix norm). It is defined by $$\|X\|=\sqrt{\operatorname{trace}(X^\ast X)}$$. In your question, just put $$X=MV_2$$ and use the implications that $$X^\ast X=0\Rightarrow\|X\|=0\Rightarrow X=0$$.

You may verify that $$\operatorname{trace}(X^\ast X)$$ is just sum of squared moduli of all entries of $$X$$. Therefore, $$\sqrt{\operatorname{trace}(X^\ast X)}$$ is identical to the Euclidean norm of the vector $$\operatorname{vec}(X)$$ obtained by stacking the columns of $$X$$ one another. E.g. if $$X=\pmatrix{1&2i\\ 3+4i&5}$$, then $$\sqrt{\operatorname{trace}(X^\ast X)}=\left\|\pmatrix{1\\ 3+4i\\ 2i\\ 5}\right\|_2.$$ So, the Frobenius norm is basically the usual Euclidean norm. But we don't view it as a vector norm (and we give it another name) because it has a property that is significant only when the norm is viewed as a matrix norm, namely, the Frobenius norm is submultiplicative: $$\|XY\|\le\|X\|\|Y\|$$.

The following is true

$$\|A\|_{2} = \sqrt{\textrm{Tr}(A^{T}A})$$

so

$$\|A\|_{2}^{2} = \textrm{Tr}(A^{T}A)$$

If we have $$A = M V_{2}$$ then the we get

$$A^{*} = (MV_{2})^{*} = V_{2}^{*}M^{*}$$

Substituting $$MV_{2}$$ for $$A$$

$$\| M V_{2}\|^{2} = \textrm{Tr}(V_{2}^{*}M^{*}M V_{2})$$

• when you write $A^T$ does this apply to the conjugate transpose? – IntegrateThis Jul 7 '19 at 0:34
• The same rule applies. I could write $\|A\|^{2} = \textrm{Tr}(A^{*}A)$ if that helps – user3417 Jul 7 '19 at 0:35