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.
Best Answer
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.