Can the cube root of this matrix be found?

Question

I tried to find the matrix X which satisfies X$^3$=$\begin{bmatrix}1 & 0\\-3 & 1\end{bmatrix}$ by using the method of diagonalization only to realise that it doesn't work for this type of matrix (which has an eigenvalue of $1$). Furthermore, I tried to use induction to show that $X^n$=$\begin{bmatrix}1 & 0\\-n & 1\end{bmatrix}$ because i figured i could just set $n=1$. I would be very grateful if someone could help. Thanks!

Answer 1

If $X = \pmatrix{a & b\cr c & d\cr}$, you want $$ \eqalign{ \left( {a}^{2}+bc \right) b+ \left( ab+bd \right) d&=0\cr \left( {a}^{2}+bc \right) a+ \left( ab+bd \right) c-1&=0\cr \left( ac+cd \right) a+ \left( bc+{d}^{2} \right) c+3&=0\cr \left( ac+cd \right) b+ \left( bc+{d}^{2} \right) d-1&=0\cr}$$ which can be solved algebraically, yielding $$ \eqalign{a &= d\cr b &= 0\cr c &= -d\cr d^3 &= 1\cr} $$ So the only real solution is indeed $$ \pmatrix{1 & 0\cr -1 & 1\cr} $$

EDIT: You may prefer this slightly more general solution. You want to solve the $2\times 2$ matrix equation $X^3 = I+N$ where $N \ne 0$, $N^2 = 0$ and $X$ is real. The eigenvalues of $X$ must all be cube roots of $1$ (the eigenvalues of $X^3$ are the cubes of the eigenvalues of $X$, by the Spectral Mapping Theorem). If the real matrix $X$ has a non-real eigenvalue, it has the complex conjugate eigenvalue as well, and then $X$ would have two distinct eigenvalues, therefore be diagonalizable, which is impossible since $I+N$ is not diagonalizable. So the only eigenvalue of $X$ is $1$ (a repeated eigenvalue). Thus $X = I + M$ where $M$ is nilpotent; since these are $2 \times 2$ matrix that means $M^2 = 0$. But then by the binomial theorem, $X^3 = (I+M)^3 = I + 3 M$, so $M=N/3$.

Answer 2

More generally, consider the equation $X^3=I_n+N$ where the unknown is $X\in M_n$ and $N\in M_n$ is a nilpotent known matrix.

Then a particular solution is given by the binomial formula

$X=(I+N)^{1/3}=I+1/3N-1/9N^2+\cdots$. Note that the previous sum is finite !