# Thread: Show that matrix is diagonal

1. ## Show that matrix is diagonal

Problem: Show that if a matrix $\textbf{A}$ is both triangular and unitary, then it is diagonal.

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My attempt at a solution:

Since $\textbf{A}$ is unitary, $\textbf{A}^*=\textbf{A}^{-1}$ so, $\textbf{A}\textbf{A}^*=\textbf{I}$.

The product of a lower-triangular matrix and a upper-triangular matrix is not the identity.
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I would appreciate it if someone could point me in the direction of a more elegant way of doing this.

Thank you.

2. We show by induction that the $j$th column of $A$ is empty except for a non-zero entry in the $j$th coordinate. Recall that $A$ is unitary if and only if its columns are an orthonormal basis of $\mathbb{C}^n$. Denote the columns of $A$ by $a_1, ..., a_n$. Then clearly $a_1$ is empty except for its first entry (because $A$ is diagonal). Now suppose this holds up to $j-1$. Then, since $=0=a_{jk}a_{kk}$ for $1 \leq k < j$, and $a_{kk} \neq 0$, we must have $a_{jk}=0$, and thus $a_j$ is also as described.

3. Originally Posted by Bruno J.
Then, since $=0=a_{jk}a_{kk}$ for $1 \leq k < j$, and $a_{kk} \neq 0$, we must have $a_{jk}=0$, and thus $a_j$ is also as described.
So $a_{kk} \neq 0$ because $\textbf{A}$ is diagonal.
Is $a_{jk}a_{kk}=0$ for $1 \leq k < j$ the same as saying that $\textbf{A}$ is diagonal?

I'm not sure how to translate $$ into English.