# Hilbert Schmidt Norm-Rank-inequality

Problem:

Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS}$ is the Hilbert Schmidt Norm.

Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.

Proof of the implication $\Leftarrow$:

• If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$
• If $rank(A)=1$, than: $A=\begin{pmatrix} A_{1}\\ \alpha _{2}A_{1}\\ ...\\ \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$

and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$

Proof of the implication $\Rightarrow$

We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$

On the other hand: $A=\begin{pmatrix} A_{1}\\ A_{2}\\ ... \\ A_{n} \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\ \left \langle A_{2},x \right \rangle\\ ...\\ \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}}$$

Where $\left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$

Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$

Please let me know if my solution makes sense.

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The rank $1$ case is not quite right, because the first row could be $0$. All the rows are multiples of some row, it's just not necessarily the first. – Robert Israel Feb 29 '12 at 22:39
Also, it's not just that the rows are linearly dependent (which would say that the rank is less than $n$), but that they are all multiples of $x$. – Robert Israel Feb 29 '12 at 22:42
@Robert Israel: If the first row is all zeros, then I can exchange the first row by a non-zero row, and I will get a different matrix $F$, but still $F$ and $A$ have the same norms . Does this make sense? – Boyan Klo Mar 1 '12 at 2:25
Yes, it does make sense. – Robert Israel Mar 1 '12 at 3:38

I guess you know the property that $\|A\|_{HS}$ is the square root of the summation of all $A$'s singular values' square and $\|A\|$ is the largest one. The rank of a matrix is the number of its nonzero singular values. With these fact, it is obvious then.