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I answered this problem how to raise a matrix to a higher power awhile ago, but I do not know if it is related. I am reading about this problem on the GRE math subject test, and it expects me to know the following given a matrix $P$:

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

I looked up the definition of a diagonizable matrix which is the following:

We say $P$ is diagonizable iff $\exists$ an invertible matrix $A$ ST $A^{-1}PA=R$ where $R$ is some matrix which has all the values zero except its diagonal.

The solution is number $37$ on page $30$ here.

$\textbf{Question:}$ Is there a more concise/efficient way to solve this problem? I have never heard of the word diagonizable anywhere in my undergraduate classes. I am having a hard time going through the solution especially knowing the time restraints are less than $3$ minutes to solve this problem.

W. G.
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  • You must have a link wrong. The link you say is to solution number 37 on page 30 goes to an MSE page. – saulspatz Aug 22 '18 at 23:09
  • Sorry about that. I changed it now. – W. G. Aug 22 '18 at 23:12
  • I can't think of a more concise argument. The important fact is that a matrix is diagonalizable if and only if it has a basis of eigenvectors. You can try Googling "Idempotent matrix is diagonalizable" to find alternate proofs. – saulspatz Aug 22 '18 at 23:20
  • Okay I'll Google that. Thank you for looking into it for me. – W. G. Aug 22 '18 at 23:24
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    Understanding a solution to this particular problem is not nearly as important as having a broad enough background in linear algebra to know the meaning of terms like "diagonalizable." – Brian Borchers Aug 22 '18 at 23:29
  • There was another answer for this, but it was taken down. It stated that you solve $P^2-P=P(P-I)=0$ where the identity acts like the number one. So, then solve for $x$ when $x(x-1)=0$ to get $x=1$ or $x=0$. As both these solutions are distinct, $P$ is diagonalizable. However, I am not sure if this is right. I have seen that $P$ is diagonizable iff the eigenvalues are all distinct. Not sure if that gave the same thing. – W. G. Aug 23 '18 at 00:53
  • Having distinct eigenvalues is sufficient but not necessary. For example, the identity matrix is obviously diagonalizable, but its only eigenvalue is $1$. On the other hand, the minimal polynomial splitting into distinct linear factors is both necessary and sufficient. – amd Aug 23 '18 at 01:21
  • Umm... what problem exactly are you asking about? Not the one that you linked to, certainly. – amd Aug 23 '18 at 01:29
  • It was posted as an answer and then taken down below. – W. G. Aug 23 '18 at 01:34

1 Answers1

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If $v\in\mathbb{R}^n$, then$$v=\bigl(v-P(v)\bigr)+P(v).$$Besides, $P\bigl(v-P(v)\bigr)=0$ and $P\bigl(P(v)\bigr)=P(v)$. It follows from this $\mathbb{R}^n$ is the direct sum of the eigenspace that corresponds to the eigenvalue $0$ with the eigenspace that corresponds to the eigenvalue $1$. So, there is a basis $\mathcal{B}=(e_1,\ldots,e_k,e_{k+1},\ldots,e_n)$ of $\mathbb R$ such that$$P(e_j)=\begin{cases}0&\text{ if }j\leqslant k\\1&\text{ otherwise.}\end{cases}$$The matrix of $P$ with respect to $\mathcal{B}$ is clearly diagonal.

José Carlos Santos
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  • I think it should be $P(P(v))=P^2(v)=P(v)$ and not just $v$ which isn't a big deal, but how do you know it follows from this that $\mathbb{R}^n=V_1\underbrace{+}_{\text{direct sum}}V_2$ where $V_1$ denotes the eigenspace corresponding to an eigenvalue of $1$ and $V_2$ denotes the eigenspace corresponding to an eigenvalue of $0$. Also, how are you getting the eigenvalues $0$ and $1$ to start? – W. G. Aug 23 '18 at 02:28
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    EIgenspaces corresponding to distinct eigenvalues always form a direct sum. This direct sum may not be the whole space, but in this case it is the whole space, because the equality $v=(v-P(v))+P(v)$ shows that $v$ can be written as the sum of an eigenvector with eigenvalue $0$ with an eigenvector with eigenvalue $1$. And I got the numbers $0$ and $1$ from the roots of the polynomial $Q(x)=x^2-x$, which is such tht $Q(P)=0$. – José Carlos Santos Aug 23 '18 at 08:18