If $A$ is normal and upper triangular then it is diagonal

Question

Let $A$ be a normal matrix in Mat$_{n\times n}(\mathbb C)$, if $A$ is upper triangular then it is diagonal

(Normal means $AA^*=A^*A$, where $A^*$ is the conjugate transpose of $A$)

If I consider the diagonal of $AA^*$, let denote $(a_{ij})=A$ and $(â_{ij})_{i,j}=AA^*$ then, since $AA^*=A^*A$

$â_{ii}=\sum\limits_{k=1}^na_{ik}\overline{a}_{ik}=\sum\limits_{k=1}^n\overline{a_{ki}}{a}_{ki}$

$\implies\sum\limits_{k=1}^n|a_{ik}|^2=\sum\limits_{k=1}^n|a_{ki}|^2$.

If I take $i=n$ then it follows that $a_{in}=0, \forall 1\le i\le n-1$ and continuing in this manner the upper diagonal entries are zero, Is this correct ?

Can I show it in another way, because in a previous exercise I had to show that ''If A is normal and nilpotent then $A=0$'' so using this can I decompose $A$ into diagonal and nilpotent matrix, then show that the nilpotent part is zero ?

Answer 1

Denote $A=\begin{pmatrix}a_{11}&a_{12}&\cdots&a_{1n}\\ &a_{22}&\cdots&a_{2n}\\&&\ddots&\vdots\\&&&a_{nn}\end{pmatrix}$. Observe that the $(1,1)$-entries of $A^\ast A$ and $AA^\ast$ are $$\vert a_{11}\vert^2\quad\mbox{and}\quad \sum_{i=1}^n\vert a_{1i}\vert^2,$$ respectively. Since $A$ is normal, $$ \sum_{i=1}^n\vert a_{1i}\vert^2=\vert a_{11}\vert^2\quad\Rightarrow\quad \sum_{i=2}^n\vert a_{1i}\vert^2=0\quad\Rightarrow\quad a_{12}=a_{13}=\cdots=a_{1n}=0.$$ Now, from preceding result, the $(2,2)$-entry of $A^\ast A$ is $$\vert a_{12}\vert^2+\vert a_{22}\vert^2=\vert a_{22}\vert^2,$$ and the $(2,2)$-entry of $AA^\ast$ is $\displaystyle\sum_{i=2}^n\vert a_{2i}\vert^2$. Again, since $A$ is normal, $$ \sum_{i=2}^n\vert a_{2i}\vert^2=\vert a_{22}\vert^2\quad\Rightarrow\quad \sum_{i=3}^n\vert a_{2i}\vert^2=0 \quad\Rightarrow\quad a_{23}=a_{24}=\cdots=a_{2n}=0.$$ Continue this process, we may conclude that the upper off-diagonal entries of $A$ are all zero. Hence, $A$ is a diagonal matrix.

Answer 2

Let's prove this by induction. Suppose $A$ is an $n\times n$ upper-triangular, normal matrix.

If $n=1$, this is trivial. If $n=2$, we have $$ A = \begin{pmatrix} a & b\\ 0 & c \end{pmatrix}$$

So \begin{align*} 0 = AA^* - A^*A &= \begin{pmatrix} a & b\\ 0 & c \end{pmatrix}\begin{pmatrix} \overline{a} & 0\\ \overline{b} & \overline{c} \end{pmatrix}- \begin{pmatrix} \overline{a} & 0\\ \overline{b} & \overline{c} \end{pmatrix}\begin{pmatrix} a & b\\ 0 & c \end{pmatrix} \newline &= \begin{pmatrix} |a|^2 + |b|^2 & b\overline{c}\\ \overline{b}c & |c|^2 \end{pmatrix} - \begin{pmatrix} |a|^2 & \overline{a}b \\ a\overline{b} & |b|^2+|c|^2 \end{pmatrix} \newline &= \begin{pmatrix} |b|^2 & b\overline{c}-\overline{a}b\\ \overline{b}c -a\overline{b} & |b|^2 \end{pmatrix}_. \end{align*}

Therefore $|b|^2 = 0$. Thus $b = 0$ and $A$ is diagonal.

Now assume the result holds for $n-1$. Then $$A = \begin{pmatrix} a & B\\ 0 & C \end{pmatrix}$$

where $B$ is an $1\times (n-1)$ matrix and $C$ is an $(n-1)\times(n-1)$ upper-triangular matrix.

Since $A$ is normal, \begin{align*} 0 = A^*A-AA^* &= \begin{pmatrix} \overline{a} & 0\\ B^* & C^* \end{pmatrix}\begin{pmatrix} a & B\\ 0 & C \end{pmatrix} - \begin{pmatrix} a & B\\ 0 & C \end{pmatrix}\begin{pmatrix} \overline{a} & 0\\ B^* & C^* \end{pmatrix} \newline &= \begin{pmatrix} |a|^2 & \overline{a}B\\ aB^* & B^*B + C^*C \end{pmatrix} - \begin{pmatrix} |a|^2 + BB^* & BC^*\\ CB^* & CC^* \end{pmatrix} \newline &= \begin{pmatrix} -|B|^2 & \overline{a}B-BC^*\\ aB^*-CB^* & B^*B + C^*C-CC^* \end{pmatrix}_. \end{align*}

Thus $|B| = 0$ and so $B=0$. Hence $C^*C-CC^* = 0$, i.e. $C$ is normal. By the hypothesis of induction, $C$ is diagonal. Finally,

$$ A = \begin{pmatrix} a & 0\\ 0 & C \end{pmatrix}_.$$ So $A$ is diagonal.