# Let A be the 2×2 integral matrix Determine for which positive integers n, there is a complex matrix B such that B^n = A.

Let $$A$$ be the 2×2 integral matrix

$$\begin{pmatrix} 1 & 1\\ -4 & 5 \end{pmatrix}$$

Determine for which positive integers $$n,$$ there is a complex matrix B such that $$B^n = A$$.

Here are my approaches.

1. Assume the matrix $$B$$ has the following form.

$$\begin{pmatrix} Z_1 & Z_2\\ Z_3 & Z_4\\ \end{pmatrix}$$

where $$Z_1,Z_2$$ are complex numbers. Hence,

$$det(B)^n=detA$$, then since the determinant is multiplicative,

$$(det(B))^n=9$$

$$(Z_1Z_4-Z_2Z_3)^n=9$$

For the moment I assumed $$n=2$$ and then yielded the following complex polynomial

$$(Z_1Z_4-Z_2Z_3)^2=9$$

I was trying to guess some complex numbers but no luck.

1. The second approach was to diagonalize the matrix $$A$$ and try to write $$B^n=P^{-1}DP$$. But it turned out that $$A$$ has repetitive eigen values and is not diagonalizable.

Could you tell me if I'm on the right track or If I'm missing something?

Thanks

• Multiple eigenvalues does not automatically imply that it is not diagonalizable.
– ameg
Commented May 29 at 18:57
• All complex invertible matrices have $n$-th roots. Commented May 29 at 18:58

Let's look at this problem in general. Let $$A$$ be an arbitrary $$2\times 2$$ complex matrix.

If $$A$$ is diagonalizable, then there exist multiple matrices $$B$$ that work for any $$n\gt 1$$. Find invertible $$P$$ such that $$PAP^{-1} = \left(\begin{array}{cc} \lambda_1 & 0\\ 0 & \lambda_2 \end{array}\right),$$ and let $$\mu_i$$ be an $$n$$th complex root of $$\lambda_i$$. Then letting $$B = P^{-1}\left(\begin{array}{cc} \mu_1 & 0\\ 0 & \mu_2 \end{array}\right)P$$ will do.

If $$A$$ is not diagonalizable, then there exists an invertible complex matrix $$P$$ and a complex number $$\lambda$$ such that $$PAP^{-1} = \left(\begin{array}{cc} \lambda & 1\\ 0 & \lambda \end{array}\right).$$ Note that $$\left(\begin{array}{cc} a & b\\ 0 & a \end{array}\right)^n = \left(\begin{array}{cc} a^n & nba^{n-1}\\ 0 & a^n \end{array}\right),$$ If $$\lambda\neq 0$$, then let $$\mu$$ be a complex $$n$$th root of $$\lambda$$, and let $$b=\frac{1}{n\lambda^{n-1}}$$. Then letting $$C = \left(\begin{array}{cc} \mu & b\\ 0 & \mu \end{array}\right)$$ gives $$C^n = PAP^{-1}$$, so letting $$B=P^{-1}CP$$ will do.

Finally, if $$\lambda = 0$$, then there is no matrix $$B$$ such that $$B^n=PAP^{-1}$$ if $$n\geq 2$$. If such a $$B$$ existed, then we would have $$B^{2n} = PA^2P^{-1} = \mathbf{0}_{2\times 2}$$; then the minimal polynomial of $$B$$ divides $$x^{2n}$$, and has degree at most two, so it equals either $$x$$ or $$x^2$$. Either way, $$B^2=\mathbf{0}_{2\times 2}$$, so $$B^n=\mathbf{0}_{2\times 2}\neq PAP^{-1}$$.

Note. For $$n=2$$, this can be done "by hand" (without relying on the minimal polynomial) as follows: letting $$B=\left(\begin{array}{cc}a&b\\c&d\end{array}\right),$$ we would have $$\left(\begin{array}{cc}0&1\\0&0 \end{array}\right) = \left(\begin{array}{cc} a&b\\c&d\end{array}\right)^2 = \left(\begin{array}{cc} a^2+bc & ab+bd\\ ac + cd & bc+d^2 \end{array}\right).$$ This would require \begin{align*} a^2+bc &= 0\\ b(a+d) &=1\\ c(a+d) &=0\\ bc+d^2 &= 0 \end{align*}. From the second and third equations we see that $$a+d\neq 0$$ and hence $$c=0$$. This means that $$a=0$$ (from the first equation), and $$d=0$$ (from the fourth equation), which makes the second equation impossible.

So $$A$$ has a complex $$2\times 2$$ matrix $$n$$th root, $$n\geq 2$$, if and only if $$A$$ is not non-diagonalizable with repeated eigenvalue $$0$$.

Your particular matrix $$A$$ has determinant $$9$$, which means it is invertible, and hence does not fall under the exception. Thus, it always has $$n$$th roots.