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

Square Roots of a Matrix Problems and Solutions

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
\[A=\begin{bmatrix}
2 & 2\\
2& 2
\end{bmatrix}.\]

 
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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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