# If $Q \in \mathbb{R}^{n \times n}$ is both upper triangular and orthogonal, then $\textbf{q}_j = \pm \textbf{e}_j, j = 1,\ldots, n$

I can get this far:

If $n = 1$, then the only matrices that are both upper triangular and orthogonal are $[1]$ and $[-1]$, so $\textbf{q}_j = \pm\textbf{e}_j, j = 1$ is true. Then if we assume that the result holds for $n = k$ and suppose that $Q$ is a $k+1 \times k+1$ matrix that is upper triangular and orthogonal, then . . . ?

It seems that to get any further, I have to use the fact that there is a $k \times k$ upper triangular, orthogonal submatrix within $Q$ that I can apply the induction hypothesis to. I can see why the $k \times k$ matrix in the upper-left or lower-right would be upper triangular, but I don't see why it would necessarily be orthogonal. It seems like it's the same as saying that every submatrix along the diagonal of a nonsingular matrix is nonsingular, but I don't think that's right. Could someone clarify this for me?

-

Hint: The inverse of an upper triangular matrix is upper triangular. The transpose of an upper triangular matrix is lower triangular. But if $A$ is orthogonal, then the transpose and the inverse are related...