Solve the equation about matrix

Question

The equation is $x^2 = x$, which $x$ is a $2\times2$ matrix. Anyone can give me some hint? Thanks!

Answer 1

Two different hints:

1) This is a question with 2-by-2 matrices... why not just write down good ol' $x=\begin{pmatrix}a & b\\c & d\end{pmatrix}$, compute $x^2-x$, and see what happens?

2) You could also notice that for such a matrix $x$, we must have $x^2-x=0$; hence the minimal polynomial for $x$ must divide $t^2-t=t(t-1)$. What does that tell you about the matrix?

Answer 2

The matrix satisfying this equation is, by definition, idempotent. A matrix will satisfy this equation if and only if it is idempotent. That is:

$$X^2=X$$

A $2\times{2}$ matrix is idempotent if and only if it is diagonalizable, and has eigenvalues $0$ and $1$.

In addition to the identity matrix $X=I_2$ and the zero matrix $X_2=O$, $X$ can therefore be constructed as follows, using any non-singular matrix $P$:

$$X=P^{-1} \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix}P$$

or

$$X=P^{-1} \begin{bmatrix} 0 & 0 \\ 0 & 1 \\ \end{bmatrix}P$$

Answer 3

$x^2 - x = x(x-1)$ has single linear factors.

Consider the Jordan Normal Form. This shows that you can split the matrix up into single blocks with eigenvalues of 0 or 1.