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.

  • 1
    $\begingroup$ Do you know about Jordan Normal form for a matrix? $\endgroup$ Commented Jun 10, 2019 at 17:38
  • $\begingroup$ @SeanHaight I do, but not in great detail. You could include it in your answer/comment if you want. $\endgroup$
    – Surzilla
    Commented Jun 10, 2019 at 17:39

2 Answers 2


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 $$X = \begin{bmatrix} \lambda^{2/3} & \frac{2}{3}\lambda^{-1/3} \\ 0 & \lambda^{2/3} \end{bmatrix}.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .