# Find All the Square Roots of a Given 2 by 2 Matrix

## Problem 513

Let $A$ be a square matrix. A matrix $B$ satisfying $B^2=A$ is call a square root of $A$.

Find all the square roots of the matrix
$A=\begin{bmatrix} 2 & 2\\ 2& 2 \end{bmatrix}.$

## Proof.

### Diagonalize $A$.

We first diagonalize the matrix $A$.
The characteristic polynomial of the matrix $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
2-t & 2\\
2& 2-t
\end{vmatrix}=t(t-4).
\end{align*}
Thus, the eigenvalues of $A$ are $0, 4$.
(Since $A$ has two distinct eigenvalues, it is diagonalizable.)

Let us find eigenvectors.
For the eigenvalue $0$, solving $A\mathbf{x}=\mathbf{0}$, we see that
$\mathbf{v}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ is an eigenvector for $0$.

For the eigenvalue $4$, solving $(A-4I)\mathbf{x}=\mathbf{0}$ yields that
$\mathbf{u}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is an eigenvector for $4$.

Thus the matrix
$S=\begin{bmatrix} \mathbf{v} & \mathbf{u} \end{bmatrix}=\begin{bmatrix} 1 & 1\\ -1& 1 \end{bmatrix}$ diagonalizes the matrix $A$, that is,
$S^{-1}AS=D,$ where $D$ is the diagonal matrix
$D=\begin{bmatrix} 0 & 0\\ 0& 4 \end{bmatrix}.$

### Determine Square Roots of $A$.

Now suppose that $B$ is a matrix such that $B^2=A$.
We have
\begin{align*}
D=S^{-1}AS=S^{-1}B^2S=(S^{-1}BS)(S^{-1}BS)=(S^{-1}BS)^2=B’^2,
\end{align*}
where we set $B’=S^{-1}BS$.

Observe that
$B’D=B’B’^2=B’^3=B’^2B’=DB’.$ Since $B’$ commutes with the diagonal matrix $D$, the matrix $B’$ is also diagonal.
(To see this directly, put $B’=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$ and compute $B’D$ and $DB’$. Then $B’D=D’B’$ requires $b=c=0$.)

Let $B’=\begin{bmatrix} a & 0\\ 0& d \end{bmatrix}$.
Since $B’^2=D$, we have
$\begin{bmatrix} a^2 & 0\\ 0& d^2 \end{bmatrix}=\begin{bmatrix} 0 & 0\\ 0& 4 \end{bmatrix},$ hence $a=0$ and $d=\pm 2$.

It follows that a square root of $A$ must be $B=SB’S^{-1}$, where $B’$ is one of
$\begin{bmatrix} 0 & 0\\ 0& 2 \end{bmatrix}, \quad \begin{bmatrix} 0 & 0\\ 0& -2 \end{bmatrix}.$

When $B’=\begin{bmatrix} 0 & 0\\ 0& 2 \end{bmatrix}$, we compute
\begin{align*}
B&=SB’S^{-1}=\begin{bmatrix}
1 & 1\\
-1& 1
\end{bmatrix}\begin{bmatrix}
0 & 0\\
0& 2
\end{bmatrix}
\frac{1}{2}\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}\6pt] &=\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix}. \end{align*} Similarly, when B’=\begin{bmatrix} 0 & 0\\ 0& -2 \end{bmatrix}, we obtain \[B=\begin{bmatrix} -1 & -1\\ -1& -1 \end{bmatrix}.

In summary, the square roots of the matrix $A$ are
$\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 & -1\\ -1& -1 \end{bmatrix}.$

## Related Question.

Problem.
Prove that a positive definite matrix has a unique positive definite square root.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

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### 1 Response

1. 07/18/2017

[…] $A$. (The less trivial question is that these are the only square roots of $A$. See the post “Find All the Square Roots of a Given 2 by 2 Matrix” […]

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##### No/Infinitely Many Square Roots of 2 by 2 Matrices

(a) Prove that the matrix $A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ does not have a square root. Namely, show...

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