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

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


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

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

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

Substituting $MV_{2}$ for $A$

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

  • $\begingroup$ when you write $A^T$ does this apply to the conjugate transpose? $\endgroup$ – IntegrateThis Jul 7 at 0:34
  • 1
    $\begingroup$ The same rule applies. I could write $\|A\|^{2} = \textrm{Tr}(A^{*}A) $ if that helps $\endgroup$ – Shogun Jul 7 at 0:35

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