# Prove that for every complex $2$x$2$ matrix $A$, there exists a matrix $X$ such that $X^3=A^2$.

This is a problem I've been stuck on for a while. I know that this statement is false if it had $$A$$ instead of $$A^2$$, for example, the matrix $$A = \begin{bmatrix} 0&1\\0&0 \end{bmatrix}$$ can't be written as $$X^3$$, but since $$A^2$$ equals $$\begin{bmatrix} 0&0\\0&0 \end{bmatrix}$$, it can be written as $$X^3$$ with $$X$$ being$$\begin{bmatrix} 0&0\\0&0 \end{bmatrix}$$ for example. Other than this, I have no insight into why this statement could be true: nothing can be assumed about diagonalizability, invertibility or anything else. I might be overlooking something simple. Thanks in advance.

• Do you know about Jordan Normal form for a matrix? Commented Jun 10, 2019 at 17:38
• @SeanHaight I do, but not in great detail. You could include it in your answer/comment if you want. Commented Jun 10, 2019 at 17:39

You already have discovered the key point: $$A^2$$ cannot be a nilpotent matrix of index $$2$$. This means $$A^2$$ is either a diagonalisable matrix, or a non-diagonalisable but invertible matrix. The former case is easy. In the latter case, you may (by a similarity transform) assume that $$A^2$$ is in Jordan form and try to find a triangular matrix cube root of $$A^2$$.

First we show that it suffices to solve the problem for Jordan Normal Forms.

Let $$A$$ be a $$2\times 2$$ matrix and let $$A = PJP^{-1}$$ for $$J$$ a Jordan Normal Form for $$A$$. If $$X^3 = J^2$$ for some matrix $$X$$ then

$$(PXP^{-1})^3 = PX^3P^{-1} = PJ^2P^{-1} = (PJP^{-1})^2 = A^2.$$

Thus it suffices to solve the problem for matrices in Jordan Normal Form.

There are two possible Jordan normal forms for a $$2 \times 2$$ matrix. We either have 2-blocks of size 1 or 1 block of size two. That is

$$J = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \text{ or } J = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}$$ where $$\lambda_1,\lambda_2,\lambda \in \mathbb{C}$$. We then have two cases we need to solve:

Case 1$$\left(J^2 = \begin{bmatrix} \lambda_1^2 & 0 \\ 0 & \lambda_2^2 \end{bmatrix}\right)$$: In this case let $$X = \begin{bmatrix} \lambda_1^{2/3} & 0 \\ 0 &\lambda_2^{2/3}\end{bmatrix}$$.

Case 2$$\left(J^2 = \begin{bmatrix} \lambda^2 & 2\lambda \\ 0 & \lambda^2\end{bmatrix}\right)$$: First quickly consider the case when $$\lambda = 0$$. Then assume $$\lambda \neq 0$$. In this case you can try a matrix of the form

$$X = \begin{bmatrix} \lambda^{2/3} & a \\ 0 & \lambda^{2/3} \end{bmatrix}$$. Then $$X^3 = \begin{bmatrix} \lambda^2 & 3a \lambda^{4/3} \\ 0 & \lambda^2 \end{bmatrix}$$ so we see we should choose $$a = \frac{2\lambda}{3\lambda^{4/3}} = \frac{2}{3}\lambda^{-1/3}$$. Thus in this case $$X = \begin{bmatrix} \lambda^{2/3} & \frac{2}{3}\lambda^{-1/3} \\ 0 & \lambda^{2/3} \end{bmatrix}.$$