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

## Problem 512

**(a)** Prove that the matrix $A=\begin{bmatrix}

0 & 1\\

0& 0

\end{bmatrix}$ does not have a square root.

Namely, show that there is no complex matrix $B$ such that $B^2=A$.

**(b)** Prove that the $2\times 2$ identity matrix $I$ has infinitely many distinct square root matrices.

Add to solve later

Sponsored Links

Contents

## Proof.

### (a) The matrix $A$ does not have a square root.

Let $B=\begin{bmatrix}

a & b\\

c& d

\end{bmatrix}$ be the matrix such that $B^2=A$.

Since

\begin{align*}

B^2=\begin{bmatrix}

a & b\\

c& d

\end{bmatrix}\begin{bmatrix}

a & b\\

c& d

\end{bmatrix}

=\begin{bmatrix}

a^2+bc & ab+bd\\

ca+dc & cb+d^2

\end{bmatrix}

=\begin{bmatrix}

a^2+bc & (a+d)b\\

(a+d)c & cb+d^2

\end{bmatrix},

\end{align*}

it follows from $B^2=A$ that

\[\begin{bmatrix}

a^2+bc & (a+d)b\\

(a+d)c & cb+d^2

\end{bmatrix}=\begin{bmatrix}

0 & 1\\

0& 0

\end{bmatrix}.\]

Comparing entries we obtain four equations

\begin{align*}

a^2+bc&=0 \tag{1}\\

(a+d)b&=1\tag{2}\\

(a+d)c&=0\tag{3}\\

cb+d^2&=0.\tag{4}

\end{align*}

Equation (3) gives $a+d=0$ or $c=0$.

If $a+d=0$, then equation (2) becomes $0=1$. This is impossible and thus $c=0$.

Since $c=0$, equations (1) and (4) yield that $a=d=0$.

However, inserting these into (2) gives $0=1$.

Hence there is no solution satisfying these equations.

Therefore, there is no square root of the matrix $A$.

### (b) The identity matrix has infinitely many square roots

Let

\[B=\begin{bmatrix}

1 & r\\

0& -1

\end{bmatrix},\]
where $r$ be an arbitrary real numbers.

Then we compute directly and obtain

\begin{align*}

B^2=\begin{bmatrix}

1 & r\\

0& -1

\end{bmatrix}\begin{bmatrix}

1 & r\\

0& -1

\end{bmatrix}=\begin{bmatrix}

1 & 0\\

0& 1

\end{bmatrix}=I.

\end{align*}

Hence $B$ is a square root of the identity matrix $I$.

Since $r$ is an arbitrary real numbers, there are infinitely many square roots of $I$.

## 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

Add to solve later

Sponsored Links