# Critique my proof that $r(A) = \sqrt{\operatorname{tr}(AA^*)}$ if and only if $A$ is normal of rank 1

I am attempting to prove that if $A$ is a non-zero complex $n \times n$ matrix, then $r(A) = \sqrt{\operatorname{tr}(AA^*)}$ if and only if $A$ is normal of rank 1, where $r(A)$ denotes the numerical radius of $A$ defined as $r(A) = \max \{ |xAx^*| \ | \ xx^* = 1 \}$ (I use row vectors here). This is my attempt thus far; the doubt I have is whether $\lVert u^*v \rVert_2 = \lVert u \rVert \lVert v \rVert$ implies $u = kv$ for some scalar $k$ (the very last paragraph of my attempt). Note that this is not the classical Cauchy-Schwarz inequality because I am using row vectors, not column vectors.

Let $\lVert A \rVert_F$ denote the Frobenius norm of $A$. If $A$ has singular values $\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_n \geq 0$. Then $$r(A) \leq \lVert A \rVert_2 = \sigma_1 \leq \sqrt{\sum_{j=1}^n \sigma_j^2} = \lVert A \rVert_F = \sqrt{\operatorname{tr}(AA^*)}.$$ In the case of equality, $A$ is given by $\sigma_1u^*v$ for some row vectors $u,v$ with $\lVert u \rVert = \lVert v \rVert = 1$. Now, observe that $$r(A) = r(\sigma_1 u^*v) = \sigma_1 r(u^*v) \leq \sigma_1 \lVert u^*v \rVert_2 \leq \sigma_1 \lVert u \rVert \lVert v \rVert = \sigma_1.$$ Since $r(A) = \sigma_1$, we have $\lVert u^*v \rVert_2 = \lVert u \rVert \lVert v \rVert$, and so $u = kv$ for some scalar $k$ satisfying $|k|=1$. We deduce that $A = k\sigma_1 u^*u$ is a normal matrix of rank 1, and the proof the complete?

• no, it's not complete since $\|u^*v\|=\|u\|\|v\|$ is true for any couple of vectors (try $u = [0,1]$ and $v=[1,0]$) – Exodd Dec 5 '17 at 15:56

$$\|u^*v\|_2^2 = \rho(v^*u u^*v) = \|u\|^2_2 \rho(v^*v) = \|u\|^2_2\|v\|^2_2$$ where $\rho$ is the spectral radius.
$\|uv^*\| = \|u\|\|v\| \iff \exists a,b\in \mathbb C: au = bv$
$$r(A)=\sigma_i \implies r(u^*v)=1 \implies \max \{ |xu^*vx^*| \ | \ xx^* = 1 \} = 1 \implies |xu^*| = |vx^*| = 1 \implies\exists a,b,c\in \mathbb C : au=bx=cv$$