Prove that a matrix $M$ is diagonalizable if $M^3 = M$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $M^2=I$.

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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:

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