$\ T $ upper triangular matrix and $\ T^*T = TT^* $ prove $\ T $ is diagonal

Question

Let $\ T $ be upper triangular matrix over $\ \mathbb C $ and I want to prove that if $\ T^*T = TT^* $ then $\ T $ must be a diagonal matrix.

My attempt:

if I set $\ a_{i,j} = [T]_{i,j} $ and $\ b_{i,j} = [T^*]_{i,j} $ then

$$\ [TT^*]_{i,j} = \sum_{j=1}^n a_{i,j} \cdot b_{j,i} = \sum_{j=1}^n a_{i,j} \cdot \overline{a_{i,j}} = \sum_{i=1}^n b_{i,j}\cdot a_{i,j} = [T^*T]_{i,j} $$

but for every $\ i > j , a_{i,j} = 0$ and for every $\ i < j , b_{i,j} = 0 $ then $\ 1 \le i \le j $ and $\ 1 \le j \le i $ then for every $\ i \not = j $ , equation can be true if both sides are. $\ 0 $ ?

I know this question probably not very complicated but I'm really getting confused when trying to work with sums of elements.

This is the same question here but I'm not sure I understand the solution.

Answer 1

The idea is correct but your definition of matrix multiplication is wrong.

We have $$(T^*T)_{ij} = \sum_{k=1}^n (T^*)_{ik}T_{kj} = \sum_{k=1}^n \overline{a_{ki}}a_{kj} = \sum_{k=1}^{\min\{i,j\}} \overline{a_{ki}}a_{kj}$$ since $a_{rs} = 0$ for $r > s$.

Similarly, $$(TT^*)_{ij} = \sum_{k=1}^n T_{ik}(T^*)_{kj} = \sum_{k=1}^n a_{ik}\overline{a_{jk}} = \sum_{k=\max\{i,j\}}^n a_{ik}\overline{a_{jk}}$$ so we have $$\sum_{k=1}^{\min\{i,j\}} \overline{a_{ki}}a_{kj} = \sum_{k=\max\{i,j\}}^n a_{ik}\overline{a_{jk}}, \quad \text{ for all }1 \le i,j \le n.$$

In particular, for $i=j=1$ we get $$|a_{11}|^2 = \sum_{k=1}^{1} \overline{a_{k1}}a_{k1} = \sum_{k=i}^n a_{1k}\overline{a_{1k}} = \sum_{k=1}^n |a_{1k}|^2$$ so $a_{12}=a_{13}=\cdots=a_{1n}=0$.

For $i=j=2$ we get $$|a_{22}|^2 = \underbrace{|a_{12}|^2}_{=0} + |a_{22}|^2 = \sum_{k=1}^{2} \overline{a_{k2}}a_{k2} = \sum_{k=2}^n a_{2k}\overline{a_{2k}} = \sum_{k=2}^n |a_{2k}|^2$$ so $a_{23}=a_{24}=\cdots=a_{2n}=0$.

Continuing inductively in this fashion we conclude $a_{ij}=0$ for all $i < j$ so $T$ is diagonal.