Prove that an upper triangular matrix $A$, such that $A^*A = AA^*$, must be diagonal. Let $A \in \mathbb{C}^{n \times n}$ be an upper triangular matrix that satisfies $A^{*}A=AA^{*}$. Prove that $A$ must be diagonal.
My attempt is to partition $A$ as follows:
$$
A = \left[\begin{array}{cc} a_{11} & \alpha\\
0 & \hat{A}\end{array}\right]
$$
where $\alpha = (a_{12}, a_{13}, \cdots , a_{1n})$ and $\hat{A}$ is $A$ with the first row and column removed. Using this partitioning, we have:
$$
A^{*}A = \left[\begin{array}{cc} a_{11}^{2} & a_{11}\alpha\\
a_{11}\alpha^{*} & \alpha^{*}\alpha + \hat{A}^{*}\hat{A}\end{array}\right]
$$
$$
AA^{*} = \left[\begin{array}{cc} a_{11}^{2} + \alpha\alpha^{*} & \alpha\hat{A}^{*}\\
\hat{A}\alpha^{*} & \hat{A}\hat{A}^{*}\end{array}\right]
$$
Examining entry (1,1) of each of these matrix products, we see that $\alpha\alpha^{*} = 0$. From this, I would like to conclude that $\alpha = 0$ and thus, the first row of $A$ has non-zero entry only at (1,1). Then repeat this process continuously on $\hat{A}$.
However, I can see a flaw in my argument. If the entries of A were real, then this argument seems like it would work. But since the entries can be complex, this means that $\alpha\alpha^{*} = 0$ even with $\alpha \ne 0$. For example, $\alpha = (1, i, 1, i)$ gives $\alpha\alpha^{*} = 0$.
Any ideas how to proceed here? I know that with this partitioning, I must have $\alpha = 0$ since the problem statement is true (i.e. $A$ is diagonal).
 A: First look at the first row $(i=1)$:
$$(AA^*)_{(1,:)}=\sum_{j} A_{1,j} \bar{A}_{j,1} = A_{1,1}\bar{A}_{1,1} + 0 + \cdots = A_{1,1}^2 = \sum_{j} \bar{A}_{j,1} A_{1,j} = (A^*A)_{(1,:)}$$
$$\iff$$
$$A_{1,j} = 0 \forall j\neq i(=1)$$
Repeat for every row, $i$.
QED
A: Partition $A$ as
$$
A = \begin{pmatrix} A_{11} & A_{12}\\0 & A_{22}\end{pmatrix}.
$$
If $A$ is normal, then
$$
AA^* = \begin{pmatrix} A_{11}A_{11}^* + A_{12}A_{12}^* & \star\\\star & \star \end{pmatrix} = \begin{pmatrix} A_{11}^*A_{11} & \star\\\star & \star \end{pmatrix} = A^*A,
$$
so $A_{11}^*A_{11} = A_{11}A_{11}^* + A_{12}A_{12}^*$. Now this means
$$
\operatorname{tr}(A_{11}^*A_{11})= \operatorname{tr}(A_{11}A_{11}^*)= \operatorname{tr}(A_{11}A_{11}^* + A_{12}A_{12}^*) = \operatorname{tr}(A_{11}A_{11}^*) + \operatorname{tr}(A_{12}A_{12}^*) \implies \operatorname{tr}(A_{12}A_{12}^*) = 0 \implies A_{12}=0.
$$
So now we have established that $A$ is block diagonal. Further, $A_{11}$ and $A_{22}$ are normal. Proceed by induction on $A_{11},A_{22}$, and further blocks to conclude that $A$ is in fact diagonal.
A: The trace of $AA^*$ is $\sum_{i,j}|a_{i,j}|^2$.
On the other hand, since $A$ commutes with $A^*$ it is normal, so unitarily diagonalizable, and the eigenvalues of $AA^*$ are the numbers $|\lambda|^2$ with $\lambda$ an eigenvalue of $A$. Now $A$ is triangular, so the egienvalues of $A$ are the numbers $a_{1,1}$, $\dots$, $a_{n,n}$, and we see that the trace of $AA^*=\sum_i|a_{i,i}|^2$.
Comparing the two descriptions of the trace of $AA^*$ we see that $A$ is diagonal.
