**Prove that a matrix that is both unitary and upper triangular must be a diagonal matrix.**

In case you don't remember what a unitary matrix is, please refer to wikipedia.

Possibly relevant theorem from the textbook:

Theorem 6.18. Let $\displaystyle \mathsf{T}$ be a linear operator on a finite-dimensional inner product space $\displaystyle \mathsf{V}$. Then the following statements are equivalent:

(a) $\displaystyle \mathsf{T}\mathsf{T}^*=\mathsf{T}^*\mathsf{T}=\mat hsf{I}$.

(b) $\displaystyle \langle\mathsf{T}(x),\mathsf{T}(y)\rangle=\langle x,y\rangle$ for all $\displaystyle x,y\in\mathsf{V}$.

(c) If $\displaystyle \beta$ is an orthonormal basis for $\displaystyle \mathsf{V}$, then $\displaystyle \mathsf{T}(\beta)$ is an orthonormal basis for $\displaystyle \mathsf{V}$.

(d) There exists an orthonormal basis $\displaystyle \beta$ for $\displaystyle \mathsf{V}$ such that $\displaystyle \mathsf{T}(\beta)$ is an orthonormal basis for $\displaystyle \mathsf{V}$.

(e) $\displaystyle \lVert\mathsf{T}(x)\rVert=\lVert x\rVert$ for all $\displaystyle x\in\mathsf{V}$.

Every time I start into this, it turns into something horribly messy. I feel like there's a trick I'm missing. Any thoughts? Or perhaps a full solution?

Thanks!