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This question relates to an argument made on pg. 14 of this paper.

We assume that $U,\hat{U} \in \mathbb{R}^{n \times m}$, with $m < n$, both have orthonormal columns. We define $\hat{U}_\perp \in \mathbb{R}^{n \times (n-m)}$ to be a matrix with orthonormal columns, all of which are orthogonal to the columns of $\hat{U}$, thus spanning the orthogonal complement of $range(\hat{U})$ in $\mathbb{R}^n$.

We then wish to show that $||\hat{U}^\top Ux||_2 = ||x||_2 \sqrt{1 - ||\hat{U}^\top_\perp U||^2_2}$.

Any explanation of how we can understand/prove this equality is greatly appreciated!

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The statement does not seem to be true. For a counterexample, consider $n=3,\,m=2$, $$ U=\pmatrix{1&0\\ 0&\tfrac1{\sqrt{2}}\\ 0&\tfrac1{\sqrt{2}}}, \ \hat{U}=\pmatrix{1&0\\ 0&1\\ 0&0}, \ \hat{U}_\perp=\pmatrix{0\\ 0\\ 1}, \ x=\pmatrix{1\\ 0}. $$ Then $\|\hat{U}^\top Ux\|_2 = \left\|\pmatrix{1\\ 0}\right\|=1$, but $\|x\|_2 \sqrt{1 - \|\hat{U}^\top_\perp U\|^2_2}=\sqrt{1-\left\|(0,\tfrac1{\sqrt{2}})\right\|_2^2}=\sqrt{\frac12}$.

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  • $\begingroup$ It should be $\sqrt{1/2}$ in the last term, I edited accordingly. $\endgroup$ Jun 22 '13 at 5:44

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