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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).

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    $\begingroup$ $A^\ast$ denotes the adjoint (aka the conjugate transpose), not the straight-up transpose. Hence, in particular, since $\alpha$ is a complex row vector, $\alpha \alpha^\ast = \|\alpha\|^2$, where $\|v\| = \sqrt{|v_1|^2+\cdots+|v_n|^2}$ denotes the usual norm on $\mathbb{C}^n$. $\endgroup$ Commented Jan 27, 2014 at 4:45
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    $\begingroup$ Note that "$*$" usually means not just transpose but conjugate transpose. So your counterexample isn't actually a counterexample. $\alpha\alpha^*$ is actually the squared norm of $\alpha$, and one of the properties of a norm is that $\|\alpha\|=0$ implies that $\alpha = 0$. So you actually have the right idea in your proof. $\endgroup$ Commented Jan 27, 2014 at 4:45
  • $\begingroup$ As the previous comment said, your proof is absolutely correct. Squirtle's prompt answer below was identical. $\endgroup$ Commented Sep 19, 2022 at 9:13

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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

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  • $\begingroup$ The equals sign at the very end is true a priori because we are assuming that $AA^*=A^*A$; this implies that the next to last equals sign must also be true. (i.e. c=e, e=f, implies c=f). $\endgroup$ Commented Jan 27, 2014 at 5:47
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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.

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    $\begingroup$ Is reaching out for the trace really needed? Your 3rd equation line yields $A_{12}\left(A_{12}\right)^* = 0$ which is satisfied (if and) only if each entry in $A_{12}$ is zero. $\endgroup$ Commented Aug 15, 2022 at 9:52
  • $\begingroup$ @Hanno You've just made me realize that I made a mistake. $\endgroup$ Commented Aug 15, 2022 at 10:00
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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.

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  • $\begingroup$ While correct, I would argue that this is not a "go to" proof, because the spectral theorem is itself proven based on the fact that a triangular normal matrix is diagonal. $\endgroup$ Commented Sep 19, 2022 at 9:37
  • $\begingroup$ @V.S.e.H. you are free to be wrong, of course :-) There is absolutely no need to base the proof of the spectral theorem on that, and that is actually very inconvenient. If $A$ is normal, then you can easily find a non-zero vector $v$ that is an eigenvector for $A$ and for $A^*$, and the orthogonal complement $v^\perp$ is invariant under $A$ and $A^*$, so you can do induction there. $\endgroup$ Commented Sep 19, 2022 at 20:41
  • $\begingroup$ It's not about being right or wrong, it's about having a self-contained proof that does not rely on other results, except the assumption that $A$ is normal, which is not the case in your proof, hence why I don't consider it a "go to" :P And, while I concede that proving the spectral theorem does not need to go trough the usual Schur triangularization step, I do find this way of proving it extremely convenient and intuitive, and it seems that this is the go to proof in most textbooks, but I guess we can agree to disagree on this part. $\endgroup$ Commented Sep 19, 2022 at 21:08
  • $\begingroup$ That proof completely obscures the fact that the diagonalizability of normal maps entirely similar to that of unitary ones — the argument I sketched is exactly the same one that one uses for unitrary matrices, for which the fact $A$ shares with $A^*$ an eigenvector is more obvious. $\endgroup$ Commented Sep 19, 2022 at 21:22

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