# $2\times 2$ matrices over $\mathbb{C}$ that satisfy $A^3=A$

Let $A$ be a $2\times 2$ matrix with complex entries. What would be the number of $2\times 2$ matrices $A$ that satisfies $A^{3} = A$. Question was are they infinite?

If it is $3\times 3$ matrix then by applying Cayley-Hamilton theorem I could have said that given matrix is diagonalizable . Also zero is eigenvalue of $A$. So it would be collection of all singular diagonalizable matrices. But how to count them?

Here I have $2\times 2$ matrices i feel i can't apply Cayley Hamiltonian theorem here? I am stuck with these thoughts?

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What about $$\begin{pmatrix}1-\frac{a^2}{a^2+b^2}&-\frac{a b}{a^2+b^2}\\-\frac{a b}{a^2+b^2}&1-\frac{b^2}{a^2+b^2}\end{pmatrix}$$ for $a,b$ real? – J. M. May 10 '12 at 3:50
Or $$\begin{pmatrix}0 & 1/x\\ x & 0\end{pmatrix}$$ – Alex Becker May 10 '12 at 3:55
what should be condition on $x$? perhaps for any non zero real $x$? – srijan May 10 '12 at 3:59
@ j.m. and , ALEXAre they any special kind of matrices? – srijan May 10 '12 at 4:00
Do you know about the minimal polynomial? – Arturo Magidin May 10 '12 at 4:01

If $A^3=A$, then $A$ satisfies the polynomial $t^3-t = t(t^2-1)=t(t-1)(t+1)$.

Since you say you know about the minimal polynomial

The minimal polynomial divides every polynomial that the matrix satisfies, so the minimal polynomial divides $t(t-1)(t+1)$. In particular, since it is squarefree, $A$ must be diagonalizable.

If the matrix has a single eigenvalue, then it is a scalar multiple of the identity; the three possibilities are $0$, $I$, and $-I$.

If the matrix has distinct eigenvalues, then the eigenvalues are either $0$ and $1$, $0$ and $-1$, or $1$ and $-1$. In any of these cases, there are infinitely many distinct matrices that satisfy these conditions. For example, for eigenvalues $0$ and $1$, any projection onto a 1-dimensional subspace will do; there are infinitely many different projections onto, say, the subspace $\{(x,0)\mid x\in\mathbb{C}\}$, one for every possible complement. That already solves the problem.

If you don't remember about the minimal polynomial, you can note that any eigenvalue $\lambda$ must be a root of the polynomial $t^3-t$ (more generally, if $A$ satisfies $f(t)$, then every eigenvalue of $A$ must be a root of $f(t)$). That leads you to the fact that $A$ has eigenvalues $0$, $1$, and $-1$, and from there you can obtain infinitely many different matrices as noted above.

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Sincere thanks to you sir – srijan May 10 '12 at 5:11

Any projection matrix satisfies $A^2=A$ and so $A^3=A$. You can get one projection matrix for each line through the origin and so there are infinitely many solutions. As Arturo has noticed, this already answers the question.

If you want to find all matrices $A$ such that $A^3=A$, you can proceed like this:

The Cayley-Hamilton theorem implies that $A^2 = \alpha A + \beta I$ for some scalars $\alpha$ and $\beta$. Then, $A=A^3= \alpha A^2 + \beta A=\alpha (\alpha A + \beta I)+ \beta A=(\alpha^2+\beta)A+\alpha\beta I$ implies $(1-\alpha^2-\beta)A=\alpha\beta I$.

If $1-\alpha^2-\beta\ne0$, we get $A=\gamma I$. In this case, we must have $\gamma^3=\gamma$ and so there are only three solutions: $A=0$, $A=I$, $A=-I$.

If $1-\alpha^2-\beta=0$, then $\alpha\beta=0$, and the only solutions come from $A^2=I$ and $A^2=\pm A$. Now there are plenty of solutions for these. Even $A^2=I$ has infinitely many solutions. See for instance Involutory matrix. Alex's comment contains an infinite subfamily.

All this is essentially Arturo's answer without using minimal polynomials or eigenvalues.

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Heartily thanks i got my answer – srijan May 10 '12 at 5:11