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How do I go about showing that an upper triangular matrix with orthogonal columns has to be a diagonal matrix?

I know that the property $M^\text{T}M = I$ should be used but I'm not sure how.

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  • $\begingroup$ Why not look at an off-diagonal component of $M^{\mathrm{T}}M = I$? $\endgroup$ – user26872 Apr 8 '12 at 23:08
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Hint: If $M$ is any invertible upper triangular matrix, then $M^{-1}$ is also upper triangular (provable via, say, Cramer's rule), while $M^t$ is lower triangular (provable via, say, direct inspection). If $M$ is orthgonal, what's the relationship between $M^t$ and $M^{-1}$?

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We write the matrix as $[v^1...v^n]$. It is orthogonal, so we have $v^1\cdot v^n=0$, and that implies $v^n_1=0$. Then we proceed by induction.

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