If $A$ is paranormal then $A$ is normal in dimension finite

Question

Let $A\in\mathcal L(V),\dim V\lt+\infty$.

We know that if $A$ normal then $A$ is paranormal. I need help to prove the following:

If $A$ is paranormal then $A$ is normal in finite-dimension.

Thank you in advance!

Answer 1

We may prove the statement by mathematical induction on $\dim V$. The base case $\dim V=1$ is trivial, because every linear operator is normal. In the inductive step, the statement follows if $A=0$. Suppose $A\ne0$. By scaling it, we may assume that its operator norm is $1$. Thus $Av=w$ for some unit vectors $v$ and $w$. As $A$ is paranormal, $$ \|Aw\|=\|A^2v\|\ge\|Av\|^2=\|w\|^2=\|w\|. $$ As $\|A\|=1$, we must have $\|Aw\|=\|w\|$. Hence $W:=\{w\in AV,\ \|Aw\|=\|w\|\}$ contains at least one nonzero vector.

It follows that $r=\dim\operatorname{span}(W)\ge1$. Let $\{Av_1,Av_2,\ldots,Av_r\}$ be a basis of $\operatorname{span}(W)$ and let $w=\sum_{i=1}^rc_iAv_i$ be an arbitrary unit vector in $\operatorname{span}(W)$. Since $A$ is paranormal, we have $$ \|Aw\|=\left\|A^2\sum_{i=1}^rc_iv_i\right\|\ge\left\|A\sum_{i=1}^rc_iv_i\right\|^2=\|w\|^2=\|w\|. $$ Again, as $\|A\|=1$, we must have $\|Aw\|=\|w\|$. Hence $W$ is a nontrivial invariant subspace of $A$. If we complete an orthonormal ordered basis of $W$ to an orthonormal ordered basis of $V$, the matrix representation of $A$ will become $$ \pmatrix{U&Y\\ 0&X} $$ where $U$ is a nonempty (because $r\ge1$) unitary matrix that represents the restriction of $A$ on $W$. Since $\|A^\ast\|=\|A\|=1$, the norm of every row of $A$ is at most $1$. Therefore $Y=0$. But then $X$ is paranormal and by induction assumption, it is normal. Therefore $A=U\oplus X$ is normal.