# $M = \text{Identity matrix} (I) - 2P_S$. Show that its eigenvalue must be $1$ or $-1$, and $M$ is diagonalizable.

$$S$$ 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.

• 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 ? Commented Oct 6, 2019 at 20:28

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.

• His explanations are always perfect. Commented Oct 6, 2019 at 21:11
• @Sebastiano: I saw that! Thanks! Cheers! Commented Oct 6, 2019 at 21:14
• how could you get D = OMO^T = OMO^(-1)? thank you Commented Oct 7, 2019 at 19:18
• @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! Commented Oct 7, 2019 at 19:34
• Thank you so much! Commented Oct 7, 2019 at 19:52

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 $$\{1,-1\}$$.

• thank you for your help! Commented Oct 7, 2019 at 19:23