Solve $X^3 = A$ in $M_2(\mathbb{R})$ where the matrix $A$ is given.

Question

Consider the matrix:

$$A = \begin{pmatrix} 3 & -2 \\ 6 & -4 \\ \end{pmatrix}$$

I have to solve the equation:

$$X^3 = A$$

where $X \in M_2(\mathbb{R})$.

First, I tried using the notation:

$$X = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$

where $a, b, c, d \in \mathbb{R}$. I raised $X$ to the third power and then equated it with $A$ hoping to get something nice. Surprise, surprise, I didn't.

Then I noticed that the determinant of $A$ is $0$ and since $X^3 = A$, that means that the determinant of $X$ is also $0$. So we have the relation:

$$ad = bc$$

in the matrix $X$. But I don't see how I could use this further or even if I should at all.

So how should I approach this exercise?

Answer 1

Hint $A$ has two eigenvalues $\lambda_1=0$ ($\because \det(A)=0$) and $\lambda_2=-1$ ($\because \text{tr}(A)=-4+3=-1 =\text{sum of eigenvalues}$).

So $A$ is diagonalizable, i.e. we can write $A=P\begin{bmatrix}0&0\\0&-1\end{bmatrix}P^{-1}=PDP^{-1}$.

Note that $\sqrt[3]{D}=D$ itself. So we can have $X=PDP^{-1}=A$.

Another way Observe that the characteristic polynomial of $A$ is $\lambda^2+\lambda=0$. Thus $A^2+A=0$. Which means $A^3=A$.

Answer 2

$A=uv^T=\pmatrix{1\\ 2}\pmatrix{3&-2}$ is a rank-one matrix. Therefore $X$ (if exists) must have rank one. It follows that $X$ and $A$ have the same column spaces and the row spaces, i.e. $X=cA$ for some scalar $c$. The equation $A=X^3$ thus reduces to $A=c^3A^3=c^3(v^Tu)^2A=c^3A$ or $c^3=1$. As $X$ is real, $c=1$. Hence $X=A$ is the only solution.