# Thread: Unitary and upper triangular?

1. ## Unitary and upper triangular?

Question: prove that a matrix that is both unitary and upper triangular must be diagonal.

Call the matrix $\displaystyle A$. By the fact that it is unitary, $\displaystyle AA^* = A^*A = I_n$. In other words, $\displaystyle \sum_{i=1}^n a_{ji} a^*_{ji} = 1$ for all $\displaystyle j=1,...,n$ where $\displaystyle a^*$ is the complex conjugate. Additionally, $\displaystyle \sum_{i=1}^n a_{ji} a^*_{ki} = 0$ for $\displaystyle j \neq k$.

And I know that $\displaystyle A*$ is lower triangular. Any hints, please? Thanks.

2. Originally Posted by Last_Singularity
Question: prove that a matrix that is both unitary and upper triangular must be diagonal.

Call the matrix $\displaystyle A$. By the fact that it is unitary, $\displaystyle AA^* = A^*A = I_n$. In other words, $\displaystyle \sum_{i=1}^n a_{ji} a^*_{ji} = 1$ for all $\displaystyle j=1,...,n$ where $\displaystyle a^*$ is the complex conjugate. Additionally, $\displaystyle \sum_{i=1}^n a_{ji} a^*_{ki} = 0$ for $\displaystyle j \neq k$.

And I know that $\displaystyle A*$ is lower triangular. Any hints, please? Thanks.

You're almost there, though your notation perhaps makes things a little messier:

$\displaystyle U=(u{_ij})$ is unitary, so:

(i) $\displaystyle \Longrightarrow\,U^{*}=U^{-1}$

(ii) $\displaystyle \sum\limits_{k=i}^n(u_{ik}\overline{u_{ik}}=)\left |u_{ik}\right|^2=1$

(iii) The inverse of an upper triangular matrix $\displaystyle U=(u_{ij})$ is also upper triangular with diagonal elements $\displaystyle =u_{ii}^{-1}=\frac{1}{u_{ii}}\,,\,i=1,2,...,n$

Now put together (i) and (iii): $\displaystyle \frac{1}{u_{ii}}=\overline{u_{ii}}\Longleftrightar row\,\left|u_{ii}\right|^2=1$ , and taking into account (ii) this gives us that all the elements outside of the main diagonal must be zero. Q.E.D.

Or simpler, using your reasoning and (iii):$\displaystyle U^{-1}$ is upper triangular, but $\displaystyle U^{-1}=U^{*}$ , and $\displaystyle U^{*}$ is LOWER triangular $\displaystyle \Longrightarrow\,U^{*}$ is both upper and lower triangular and then it is diagonal, and thus also $\displaystyle U$ is.
Tonio