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Let $M$ be a real square matrix such that $M^3 = M$. Prove that $M$ is diagonalizable.

Proof:

We have a $M$, a $2\times2$ matrix, $M$ is diagonalizable if $M$ has 2 linearly independent eigenvectors ${v_1,v_2}$, if so, then: $D = P^{-1}MP$ We give a concrete example:

Let $$ M = \begin{bmatrix}1&3\\2&2\end{bmatrix} $$ then the eigenvalues $x_1 =-1$ and $x_2 = 4$ with eigenvectors $(3,-2)$ and $(1,1)$, respectively.

Hence, $$ P = \begin{bmatrix}3&1\\-2&1\end{bmatrix}, D = \begin{bmatrix}-1&0\\0&4\end{bmatrix} $$

This is my proof but I am not sure about the given condition in the statement $M^3 = M$, how can I use it?

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    $\begingroup$ Hint : The polynomial $x^3-x$ does not have multiple roots. $\endgroup$
    – Peter
    Apr 2, 2018 at 14:36
  • $\begingroup$ The matrix is diagonalizable if its minimal polynomial and characteristic polynomial are the same. $\endgroup$
    – shrimpabcdefg
    Apr 2, 2018 at 14:36
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    $\begingroup$ Your "proof" has nothing whatever to do with the problem - it's just an example of a diagonalizable matrix. $\endgroup$
    – David C. Ullrich
    Apr 2, 2018 at 14:40
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    $\begingroup$ @Peter It would be more clear to say "does not have repeated roots." I was confused as $x^3-x$ certainly has more than one root! $\endgroup$
    – Math1000
    Apr 2, 2018 at 14:57
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    $\begingroup$ Sorry, I did not have this interpretation in mind. I would have formulated "more than one root" in this case , as you did , if I would have meant that. $\endgroup$
    – Peter
    Apr 2, 2018 at 14:59

3 Answers 3

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Just because, here's a one line proof: $$ x = (I - M^2)x + \frac 12 (M^2 + M)x + \frac 12 (M^2 - M)x $$ The first term is an eigenvector associated with $0$. The second is an eigenvector associated with $1$. The third is an eigenvector associated with $-1$. Hence, every vector can be written as the sum of eigenvectors.

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  • $\begingroup$ Peter's hint points towards the more typical approach here. $\endgroup$
    – Ben Grossmann
    Apr 2, 2018 at 14:46
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    $\begingroup$ It would be helpful if you have indicated how you have arrived at this one-line proof $\endgroup$
    – Wojowu
    Apr 2, 2018 at 14:47
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    $\begingroup$ @Wojowu the only helpful comment I can make is that there are analogous (but better known) proofs for the cases of $M^2 = M$ and $M^2 = I$. $\endgroup$
    – Ben Grossmann
    Apr 2, 2018 at 14:51
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    $\begingroup$ @Wojowu I was going to post the same argument (including an explanation of why those things are in fact eigenvectors). Where it comes from: $\Bbb R[x]$ is a PID and $x^2-1$, $x^2-x$ and $x^2+x$ have no common factor... $\endgroup$
    – David C. Ullrich
    Apr 2, 2018 at 15:03
  • $\begingroup$ The idea is that the polynomials $x$, $x-1$, and $x+1$ are pairwise relatively prime, so the pairwise products $(x-1)(x+1)$, $x(x+1)$, $x(x-1)$ are relatively prime. Therefore, we can get a combination $a(1-x)^2 + b x(x+1) + c x(x+1)=1$. Apriori $a$,$b$, $c$ are polynomials in $x$ with rational coefficients, but in this case they can be chosen to be constants. Note that the operators thus obtained $I-M^2$, $\ldots$ are projectors. $\endgroup$
    – orangeskid
    Apr 2, 2018 at 15:05
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I'll synthesize the comments into an answer, and add a remark.

  1. A matrix can be diagonalized over a field if and only if the minimal polynomial of the matrix has no repeated roots and splits completely (reason this by looking at the Jordan canonical form).

  2. Given (1), consider $x^3-x = x(x-1)(x+1)$, which has no repeated roots unless...

  3. The field has characteristic two.

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We have two cases:

  1. If inverse does not exist, then at least one eigenvalue $\lambda_1=0$. The second eigenvalue must therefore be real (why?). If $\lambda_2=0$ then $M$ nilpotent which can't be true since $M^2=I$ would be impossible and if not $\lambda_2 = 0$ then we have two distinct eigenvalues and then we are done because we are sure it must be diagonalizable.
  2. if $M^{-1}$ exists, then $M^{-1} = M$ (by multiplying $M^{-2}$). A real matrix being it's own inverse. What can we say about it's eigenvalues? They must be a complex conjugate pair or real (because any matrix of real entries has that property). If real then $\pm 1$ is only possibility. Complex conjugate pair is out of the question since both must fulfill squaring equalling $1$ since we have $M^2 = I$.

So the only possible ways to avoid diagonalizability is to have $2\times 2$ Jordan blocks of repeated eigenvalues $1$ or $-1$. But neither block would square to $I$ which would be required for $M^2=I$.

For the curious student these Jordan-blocks will in fact square like this:

$$\left[\begin{array}{cc}1&1\\0&1\end{array}\right]^2 = \left[\begin{array}{cc}1&2\\0&1\end{array}\right]$$

$$ \left[\begin{array}{rr}-1&1\\0&-1\end{array}\right]^2= \left[\begin{array}{rr}1&-2\\0&1\end{array}\right]$$

These blocks are of fundamental importance as it turns matrix multiplication into addition and subtraction. In general:

$$\left[\begin{array}{cc}1&a\\0&1\end{array}\right] \left[\begin{array}{cc}1&b\\0&1\end{array}\right] = \left[\begin{array}{cc}1&a+b\\0&1\end{array}\right]$$

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