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$\textbf{Definition:}$ Let $P$ be a matrix. We say is diagonalizable matrix iff

$\exists$ an invertible matrix $A$: $\exists$ a diagonal matrix $D$: $A^{-1}PA=D$.

I recently asked a question as to why the following proposition holds true:

$\textbf{Proposition:}$ Let $P$ be a $n\times n$ matrix.

If $P^2=P$, then $P$ is diagonalizable.

$\textbf{Question:}$ Let $v\in \mathbb{R}^n$. I am trying to better understand this proposition (see below):

$\textbf{Why}$ do we rewrite $v=v-Pv+Pv$? It strikes me as an obvious statement which holds true, but it is always rewritten for a particular reason. And $\textbf{what}$ is the general idea as to why eigenvalues are roots of the following expression, $Q(P):=P^2-P$? I thought the eigenvalues might have been associated with the Cayley-Hamilton theorem the way things are set up, but I could be wrong.

Roots for $Q(P)$,

<p>$P^2-P=0\implies P(P-1)=0\implies 1 \text{ and } 0 \text{ are Eigenvalues}$.</p>
W. G.
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    https://en.wikipedia.org/wiki/Projection_(linear_algebra) – Giuseppe Negro Aug 23 '18 at 12:36
  • I did not know that it was a projection. How does it being projection explain the why and what part of the question? I did find that helpful though. – W. G. Aug 23 '18 at 12:41
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    If $P$ is a projection then $I-P$ is the complementary projection (it is explained in the Wikipedia page). This at least clarifies why one makes $I-P$ appear. A little more thought also reveals why the eigenvalues are $0$ and $1$. – Giuseppe Negro Aug 23 '18 at 12:43
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    (I disagree with the downvote and the close vote). – Giuseppe Negro Aug 23 '18 at 12:47

2 Answers2

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I'll try to give some more details as an answer.

From the relation $\,P^2=P$, you obtain that $\;P(I-P)=0$, so $\operatorname{Im}(I-P)\subset \ker P$, and conversely, you can check $\ker P\subset \operatorname{Im}(I-P)$. Thus $\ker P=\operatorname{Im}(I-P)$, and if $P\ne I$, this image is non-zero so you have the eigenspace for the eigenvalue $0$.

Now $P^2-P$ corresponds to the polynomial $X^2-X$, which is the minimal polynomial of $P$ if $P\ne 0,I$. As the eigenvalues are the roots of the minimal polynomial, you know there are no other eigenvalue than $0$ and $1$.

Further, $\operatorname{Im}(P)=\ker(I-P)$ is clearly the eigenspace for the eigenvalue $1$.

Bernard
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  • Why is $ker(P)\subseteq Im(I-P)$? I tried the following: Let $y\in ker(P)$. So, $P(y)=\vec{0}$. [Show $y\in Im(I-P)$. In other words, show $(I-P)z=y$ for some $z\in \mathbb{R}^n$. ] But then I am not sure how to show that knowing $P(I-P)=0_{n\times n}$. – W. G. Aug 23 '18 at 15:08
  • @W.G. What is $(I-P)(y)$? – N. S. Aug 23 '18 at 15:28
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    For the last point, just expand by distributivity: $P(I-P)=P-P^2=0$. For the inclusion of the kernel in the image, if $Pv=0$, you can write $v=v-Pv=(I-P)v$. – Bernard Aug 23 '18 at 15:29
  • I think you meant $(I-P)z$. Using $(I-P)y$, I would say it's the [identity matrix minus $P$] multiplied by the matrix $y$. – W. G. Aug 23 '18 at 15:30
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    @W.G. NO, I mean $y$. $(I-P)(y)=y-P(y)=y$ since $y \in \ker(P)$. Since $y=(I-P)(y)$ you get that $y \in Im(I-P)$. – N. S. Aug 23 '18 at 15:32
  • @N.S. Lol on the NO. I get exactly what you are saying. I appreciate your comment! – W. G. Aug 23 '18 at 15:37
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Let $P$ be so that $P^2=P$. Let $z \in \mathbb R^n$. Define $$z_1=P(z) \\ z_2=z-z_1$$

Then, $$P(z_1)=z_1=1 \cdot z_1 \\ P(z_2)=0=0\cdot z_1\\ z=z_1+z_2$$

This shows that each vector can be written as linear combination of eigenvectors for $P$. Deduce from here that, if $E_0,E_1$ are the eigenspaces, then $$\dim(E_1)+\dim(E_0)=n$$ and hence $P$ is diagonalizable.

N. S.
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