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$S$ a subspace of $\Bbb R^n$, $P_S$ is the matrix of orthogonal projection onto $S$. $M = \text{Identity matrix}(I) - 2P_S$. Show that if $\lambda \in \Bbb R$ is an eigenvalue of $M$, then $\lambda = \pm 1$ and $M$ is diagonalizable.

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    $\begingroup$ This is the raw text of your exercise. Attempt to give a little more : begin a dialog with us... What is your work on this subject ? $\endgroup$
    – Jean Marie
    Commented Oct 6, 2019 at 20:28

2 Answers 2

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Since $P_S$ is an orthogonal projection, by definition

$P_S^T = P_S = P_S^2; \tag 1$

then

$M = I - 2P_S \tag 2$

is a symmetric matrix, for

$M^T = I^T - 2P_S^T = I - 2P_S = M; \tag 3$

thus $M$ may be diagonalized by some orthogonal matrix $O$:

$D = OMO^T = OMO^{-1}, \tag 4$

where

$O^TO = OO^T = I, \tag 5$

so that

$O^{-1} = O^T; \tag 6$

the diagonal entries of $D$ are the eigenvalues of $M$; furthermore,

$D^2 = OMO^T OMO^T = OMIMO^T$ $= OM^2O^T = OIO^T = OO^T = I; \tag 7$

therefore the diagonal entries $\mu_j$ of $D$, which are the eigenvalues of $M$, all satisfy

$\mu_j^2 = 1; \tag 8$

hence every

$\mu_j = \pm 1. \tag 9$

Note Added in Edit, Monday 7 October 2019 12:36 PM PST: It has been called to my attention in the comment of to this answer by Qazwsx199 that I have neglected to show why

$M^2 = I, \tag{10}$

where $M$ is as in (2); observe, using (1):

$M^2 = (I - 2P_S)^2$ $= I - 4P_S + 4P_S^2 = I - 4P_S + 4P_S = I. \tag{11}$

End of Note.

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    $\begingroup$ His explanations are always perfect. $\endgroup$
    – Sebastiano
    Commented Oct 6, 2019 at 21:11
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    $\begingroup$ @Sebastiano: I saw that! Thanks! Cheers! $\endgroup$ Commented Oct 6, 2019 at 21:14
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    $\begingroup$ how could you get D = OMO^T = OMO^(-1)? thank you $\endgroup$
    – Qazwsx199
    Commented Oct 7, 2019 at 19:18
  • $\begingroup$ @Qazwsx199: this is from the spectral theorem of symmetric matrices; every symmatric real matrix may be diagonalized by an orthogonal matrix; google about and you'll find out more. Check it in wikipedia. Cheers! $\endgroup$ Commented Oct 7, 2019 at 19:34
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    $\begingroup$ Thank you so much! $\endgroup$
    – Qazwsx199
    Commented Oct 7, 2019 at 19:52
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Since a projection matrix $P$ satisfies $P^2=P$ by definition, all its eigenvalues must satisfy $\lambda\in\{0,1\}$. Thus $-2P$ has eigenvalues in $\{0,-2\}$. Finally, adding the identity matrix to $-2P$ (or any matrix in general) has the effect of shifting the eigenvalues up by $1$, so $I-2P$ has eigenvalues in $\{1,-1\}$.

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  • $\begingroup$ thank you for your help! $\endgroup$
    – Qazwsx199
    Commented Oct 7, 2019 at 19:23

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