If $T \in \mathbb{C}^{n \times n}$ is an upper triangular matrix, and $TT^{H}=T^{H}T$, where $T^H$ means the Hermitian transpose of $T$, show that $T$ has to be a diagonal matrix.

What I Have Done

This seems to be obvious, but writing $T = [t_1, t_2, \cdots, t_n]$ does not help. I also tried to write $TT^{H}$ in an element-wise form, but I got stuck halfway.

Could anyone help me, thank you in advance.

  • $\begingroup$ What is the meaning of $T^H$? $\endgroup$ – José Carlos Santos Nov 26 '17 at 20:15

Let’s use induction on dimension. The claim is vacuously true for $n=1$.

If $n\geq 2$, then partition the upper-triangular matrix $T$ as follows:

$$T=\left[\begin{array}{c}S&x\\0&c\end{array}\right]$$ where $S\in\mathbb C^{(n-1)\times(n-1)}$, $x\in\mathbb C^{(n-1)\times 1}$, $0\in\mathbb C^{1\times (n-1)}$, and $c\in\mathbb C$. Then, we have that $$T^H=\left[\begin{array}{c}S^H&0\\x^H&\overline c\end{array}\right]$$ where the bar denotes complex conjugate. It is easy to compute that \begin{align*}TT^H=&\,\left[\begin{array}{c}SS^H+xx^H&\overline cx\\cx^H&|c|^2\end{array}\right]\\T^HT=&\,\left[\begin{array}{c}S^HS&S^Hx\\x^HS&x^Hx+|c|^2\end{array}\right]\end{align*} If $T T^H=T^HT$, then one must have, in particular (look at the lower right corners), that $x^Hx+|c|^2=|c|^2$, which implies that $x=0$.

It follows (look at the upper left corners) that $SS^H=S^HS$, which, by the induction hypothesis and the fact that $S$ is upper-triangular, implies that $S$ is a diagonal matrix. Since we also have $x=0$, $T$ is a diagonal matrix, too, so that the induction goes through.

  • $\begingroup$ I made a mistake when I asked the question, I have corrected the question. $\endgroup$ – Mr.Robot Nov 26 '17 at 21:25
  • $\begingroup$ @Mr.Robot I see, answer edited. $\endgroup$ – triple_sec Nov 26 '17 at 21:48

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