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In this question, someone proves that singular values of projection matrix must larger than 1:

A projection $P$ is orthogonal if and only if its spectral norm is 1 enter image description here

Is it a right rule of singular values of projection matrix? (I know that the singular values of "orthogonal projection" must be 1 or 0)

Chris
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    "I cannot find the example of projection matrix that has singular value <= 1." Try: $\left[\begin{matrix}1 & 0\0 & 0\end{matrix}\right]$ – user8675309 Mar 18 '22 at 19:01
  • Oops, I mean, the singular values are either 0 or <= 1. I edited my question. Thank you! – Chris Mar 18 '22 at 21:55
  • Your edits don't make sense to me. Anyway the result in that link is correct. – user8675309 Mar 18 '22 at 21:57
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    @Chris Based on the comments and your edits, you seem to have missed the fact that $1 \leq 1$. – Ben Grossmann Mar 19 '22 at 00:01
  • As a side remark, the problem statement in the linked question is, strictly speaking, wrong, because the zero matrix is technically also an orthogonal projection. – user1551 Mar 19 '22 at 02:46
  • Sorry for my rambling question. I edited again. I'm not saying the whole question, but the comment in the question. My true question was that I think the proof of the comment was wrong (Pu_k = u_k implies s_k v_k ^T u_k = 1) and I want to find the projection matrix who has at least one 0 < singular value < 1 but I can't. – Chris Mar 19 '22 at 12:50
  • Singular values are by definition the eigenvalues of $(P^*P)^{1/2}$ or $(P^TP)^{1/2}$ in the real case. The largest singular eigenvalue is equal to $|P|.$ When $P$ is a non orthogonal projection then $|P|>1..$ – Ryszard Szwarc Mar 19 '22 at 19:33

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Let $P$ is a projector which directly implies $P^2 = P$. Let denote the SVD of the P as follows. $U\Sigma V^T = P $. The SVD results also can be represented as follows. $P = \sum_{i} s_{i}u_{i}v_{i}^T$.

If we project a vector on the range of $P$, we will not see any difference between the original vector and the projected one. In other words, for the $u_{k}$ (note that all $u_{k}$ is in the range of $P$), we have $$Pu_{k} = u_{k}$$

Then we also have $$\sum_{i} s_{i}u_{i}v_{i}^Tu_{k} = u_{k}$$

Note that $s_{i}$ and $v_{i}^Tu_{k}$ are scalars, we have $$\sum_{i} (s_{i}v_{i}^Tu_{k})u_{i} = u_{k}$$

Note that the left singular vectors are linearly independent with each other. It means that, there is a only way to make the above equation correct that selects $ s_{k}$ and $v_{k}^Tu_{k}$ such that $(s_{k}v_{k}^Tu_{k})=1$. Otherwise, we cannot make the equality as true, because $u_{k}$ cannot be represented by a linear sum of $u_{i}$s where $i\neq k$.

StandTall
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