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Let $A$ be a real, upper-triangular, $n\times n$ matrix that commutes with its transpose. How can I show that $A$ is diagonal?

I want to show that $A-A^t =0$ but I can't. Please help me.

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Hint: One approach applies the following observation: if $A$ is block upper-triangular with $$ A = \pmatrix{B & C\\0 & D} $$ Then $$ A^TA = \pmatrix{B^TB & B^TC\\C^TB & C^TC + D^TD}\\ AA^T = \pmatrix{BB^T + CC^T & CD^T\\D^TC & DD^T} $$ By taking the trace of each side, show that $$ C^TC + D^TD = DD^T \implies C = 0 $$

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  • $\begingroup$ If $C$ is 0 then how $A$ is diagonal? $\endgroup$ – user559200 Jul 26 '18 at 19:36
  • $\begingroup$ The point is that $C$ is $0$ in every possible partition $\endgroup$ – Omnomnomnom Jul 26 '18 at 22:51
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$A^TA = AA^T$ means that the matrix $A$ is normal. This implies $\|Ax\| = \|A^Tx\|, \forall x\in\mathbb{R}^n$.

We have

$$|a_{11}|^2 = \|\text{first column of } A\|^2 = \|Ae_1\|^2 = \|A^Te_1\|^2 = \|\text{first row of } A\|^2 = \sum_{j=1}^n |a_{1j}|^2$$

so it follows that $a_{12} = a_{13} = \cdots = a_{1n} = 0$.

Continuing we get

$$|a_{22}|^2 = \|\text{second column of } A\|^2 = \|Ae_2\|^2 = \|A^Te_2\|^2 = \|\text{second row of } A\|^2 = \sum_{j=2}^n |a_{2j}|^2$$

so it follows that $a_{23} = a_{24} = \cdots = a_{2n} = 0$.

Continuing inductively we get that $A$ is diagonal.

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