A complex square ($n\times n$) matrix $A$ is called normal if
\[A^* A=A A^*,\] where $A^*$ denotes the conjugate transpose of $A$, that is $A^*=\bar{A}^{\trans}$.
A matrix $A$ is said to be nilpotent if there exists a positive integer $k$ such that $A^k$ is the zero matrix.

(a) Prove that if $A$ is both normal and nilpotent, then $A$ is the zero matrix.
You may use the fact that every normal matrix is diagonalizable.

(b) Give a proof of (a) without referring to eigenvalues and diagonalization.

(c) Let $A, B$ be $n\times n$ complex matrices. Prove that if $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$, where $I$ is the $n\times n$ identity matrix.

Proof.

(a) If $A$ is normal and nilpotent, then $A=O$

Since $A$ is normal, it is diagonalizable. Thus there exists an invertible matrix $P$ such that $P^{-1}AP=D$, where $D$ is a diagonal matrix whose diagonal entries are eigenvalues of $A$.

Since $A$ is nilpotent, all the eigenvalues of $A$ are $0$. (See the post “Nilpotent matrix and eigenvalues of the matrix” for the proof.)
Hence the diagonal entries of $D$ are zero, and we have $D=O$, the zero matrix.

It follows that we have
\begin{align*}
A=PDP^{-1}=POP^{-1}=O.
\end{align*}
Therefore, every normal nilpotent matrix must be a zero matrix.

(b) Give a proof of (a) without referring to eigenvalues and diagonalization.

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$.
We prove by induction on $k$ that $A=O$.
The base case $k=1$ is trivial.

Suppose $k>1$ and the case $k-1$ holds. Let $B=A^{k-1}$. Note that since $A$ is normal, the matrix $B$ is also normal.
For any vector $x \in \C^n$, we compute the length of the vector $B^*Bx$ as follows.
\begin{align*}
&\|B^*Bx\|=(B^*Bx)^*(B^*Bx) && \text{by definition of the length}\\
&=x^*B^*(BB^*)Bx\\
&=x^*B^*(B^*B)Bx && \text{since $B$ is normal}\\
&=x^* (B^*)^2B^2x,
\end{align*}
and the last expression is $O$ since $B^2=A^{2k-2}=O$ as $k \geq 2$ implies $2k-2 \geq k$.
Hence we have $B^*Bx=\mathbf{0}$ for every $x\in \C^n$.

This yields that
\begin{align*}
\|Bx\|&=(Bx)^*(Bx)\\
&=x^*B^*Bx=0,
\end{align*}
for every $x\in \C^n$, and hence $B=O$.

By the induction hypothesis, $A^{k-1}=O$ implies $A=O$, and the induction is completed.
So the matrix $A$ must be the zero matrix.

(c) If $A$ is normal and $B$ is nilpotent such that $A+B=I$, then $A=I$

We claim that the matrix $B$ is normal as well. If this claim is proved, then part (a) yields that $B=O$ since $B$ is a nilpotent normal matrix, which implies the result $A=I$.

To prove the claim, we compute
\begin{align*}
B^* B&=(I-A)^* (I-A)\\
&=(I-A^*)(I-A)\\
&=I-A-A^*+A^*A,
\end{align*}
and
\begin{align*}
B B^*&=(I-A) (I-A)^*\\
&=(I-A)(I-A^*)\\
&=I-A^*-A+AA^*\\
&=I-A^*-A+A^*A \qquad \text{ since $A$ is normal.}
\end{align*}
It follows that we have $B^* B=BB^*$, and thus $B$ is normal.
Hence, the claim is proved.

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