# 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

[…] $A$. (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” […]