Finding a set of similarity representatives for a collection of $2\times 2$ complex matrices

Here is the problem statement:

Find a subset $Y$ of $X:=\{A \in \text{Mat}_{2\times 2}(\mathbb{C})\ |\ A^4=A\}$ so that the following occur:

1. If $A$ $\in$ $X$ then $\exists$ $B$ $\in$ $Y$ such that $A$ and $B$ are similar.
2. If $A$ and $B$ $\in$ $Y$ and $A \ne B$, then they are not similar.

My attempt of solving this problem: If $A$ and $B$ are similar, then this means that they have the same eigenvalues and they also have the same characteristic equation. So essentially we want to find a set, so that within this set every $A$ and $B$ don't have the same eigenvalues, but with every element of $X$ the matrices have the same eigenvalues.

Ok, this is where I get confused, why do we need $A^4=A$?

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Hint: What are the possible Jordan canonical forms of $2\times 2$ matrices with $A^4=A$?

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Ok but what does being complex have to do with it – bubbly Feb 29 '12 at 20:06
@ann: Matrices over the complex numbers always have Jordan canonical forms; matrices over other fields do not (what if the characteristic polynomial does not split?) – Arturo Magidin Feb 29 '12 at 20:14
If the field was $\mathbb R$, the characteristic polynomial would factor as $\lambda (\lambda - 1)(\lambda^2 + \lambda + 1)$; you couldn't have one non-real eigenvalue without its complex conjugate, but you could have both, e.g. $\pmatrix{0 & 1 \cr -1 & -1\cr}$ – Robert Israel Feb 29 '12 at 22:08

You are making a serious logical error in your attempt.

You are correct that if $A$ and $B$ are similar, then they have the same characteristic equation (and the same eigenvalues).

You are concluding from this that if $A$ and $B$ are not similar, then they have different eigenvalues. That is not true; it is a classic logical fallacy called denying the antecedent. This is not true in general. It happens to be true for elements of $X$, but it needs to be justified, it cannot simply be asserted.

Here are two approaches: high-tech (if you know about Jordan canonical forms, Cayley-Hamilton, minimal polynomials, etc), and low-tech (if you don't know about Jordan canonical forms, Cayley-Hamilton, minimal polynomials).

High-tech. Since $A^4 = A$, then $A$ satisfies the polynomial $x^4-x = x(x^3-1) = x(x-1)(x-\omega)(x-\omega^2)$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is a primitive cubic root of unity. Therefore, the minimal polynomial of $A$ must be a degree $1$ or degree $2$ factor of this polynomial. Since the minimal polynomial has no repeated factors, the matrix $A$ must be diagonalizable. Therefore, all matrices in $X$ are similar to diagonal matrices, and two diagonal matrices are similar if and only if they have identical diagonal matrices, except perhaps for the order (Now the claim that non-similar matrices in $X$ must have a distinct set of eigenvalues is justified; but we needed to use the fact that matrices in $X$ are diagonalizable, which requires us to use the fact that they satisfy $A^4=A$; since you explicitly say you did not use it, you cannot have justified the claim about distinct eigenvalues appropriately). This suggests that we should take our set $Y$ to consist of diagonal matrices, and simply find all possible diagonal matrices that satisfy $A^4=A$.

Low-tech. If $\lambda$ is an eigenvalue of $A$, then there exists $\mathbf{x}\neq\mathbf{0}$ such that $A\mathbf{x}=\lambda \mathbf{x}$. Since $A^n\mathbf{x}=\lambda^n\mathbf{x}$, then from $A^4=A$ we conclude that $\lambda^4=\lambda$, so $\lambda$ is a root of $x^4-x = x(x-1)(x-\omega)(x-\omega^2)$. So the only possible eigenvalues of $A$ are $0$, $1$, $\omega$, and $\omega^2$.

One possibility is that $A$ is diagonalizable; there are four scalar multiples of the identity that are diagonal and have eigenvalues among the given ones; and there are $\binom{4}{2}$ diagonal matrices (up to the order of the diagonal entries) that have two distinct eigenvalues. This gives us our beginning of the set $B$.

What about the remaining possibility? If $A\in X$ is not diagonalizable, then it must have repeated eigenvalues; there is at least one eigenvector for $A$, and by extending that to some basis we obtain a basis relative to which the matrix of $A$ is of the form $$\left(\begin{array}{cc} \lambda & r\\ 0 & s \end{array}\right).$$ Since the characteristic polynomial of such a matrix is $(x-\lambda)(x-s)$, we must have $\lambda = s$. So $A$ is similar to a matrix of the form $$\left(\begin{array}{cc} \lambda & r\\ 0 & \lambda \end{array}\right)$$ for some $r\neq 0$, with $\lambda\in \{0,1,\omega,\omega^2\}$. Now, since $A^4=A$, then computing we will obtain a condition on $r$ in terms of $\lambda$ that must be satisied; using the fact that this condition must be satisfied and that $\lambda^4 = \lambda$ you can determine the possible values for $r$ and go from there.

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