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,



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


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?


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

1 Answer 1


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.


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