# A Square Root Matrix of a Symmetric Matrix

## Problem 59

Answer the following two questions with justification.

**(a)** Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$? Here $O$ denotes the $2 \times 2$ zero matrix.

**(b) **Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ where

\[A=\begin{bmatrix}

1 & -1 & 0 \\

-1 &2 &-1 \\

0 & -1 & 1

\end{bmatrix}\,\,\,\,?\]

(*Princeton University Linear Algebra Exam*)

## Hint.

- (a) Consider the eigenvalues and the Jordan canonical form of $A$
- (b) Diagonalize the matrix $A$

## Solution.

### (a) Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$?

We claim that there is no such $2 \times 2$ matrix $A$.

Suppose that we have $A^3=O$. This implies that all the eigenvalues of $A$ are zero.

To see this, let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be a corresponding eigenvector. Then applying the defining relation $A\mathbf{x}=\lambda \mathbf{x}$ successively, we have

\[\mathbf{0}=A^3\mathbf{x}=\lambda A^2\mathbf{x}=\lambda^2 A\mathbf{x}=\lambda^3 \mathbf{x}.\]
Thus, the eigenvalue $\lambda$ must be zero since $\mathbf{x}$ is a nonzero vector since it is an eigenvector.

Therefore, there exists an invertible matrix $P$ such that the Jordan canonical form of $A$ is $P^{-1}AP=\begin{bmatrix}

0 & a\\

0& 0

\end{bmatrix}$ for some number $a$.

Then $A=P\begin{bmatrix}

0 & a\\

0& 0

\end{bmatrix} P^{-1}$ and we have

\begin{align*}

A^2&=\left( P\begin{bmatrix}

0 & a\\

0& 0

\end{bmatrix} P^{-1}\right) \left(P\begin{bmatrix}

0 & a\\

0& 0

\end{bmatrix} P^{-1}\right)= P\begin{bmatrix}

0 & a\\

0& 0

\end{bmatrix}^2P^{-1} \\[6pt]
&=P\begin{bmatrix}

0 & 0\\

0& 0

\end{bmatrix}P^{-1}=O.

\end{align*}

Hence $A^2$ must be zero, and we conclude that there is no $2 \times 2$ matrix $A$ satisfying both $A^3=O$ and $A^2 \neq O$.

### (b) Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$?

The answer is yes and we will construct such a matrix $B$ as follows.

The characteristic polynomial $p(t)$ of the matrix $A$ is

\[p(t):=\det(A-tI)=-t(t-1)(t-3).\]
Thus the eigenvalues are $\lambda=0, 1, 3$.

The corresponding eigenvalues are obtained by solving the system $(A-\lambda I)\mathbf{x}=\mathbf{0}$ and we see that

\[\begin{bmatrix}

1 \\

1 \\

1

\end{bmatrix}, \begin{bmatrix}

1 \\

0 \\

-1

\end{bmatrix}, \begin{bmatrix}

1 \\

-2 \\

1

\end{bmatrix}\]
are eigenvectors corresponding to eigenvalues $0,1,3$, respectively.

Thus the matrix

\[P=\begin{bmatrix}

1 & 1 & 1 \\

1 &0 &-2 \\

1 & -1 & 1

\end{bmatrix}\]
is invertible and diagonalize the matrix $A$ such that

\[P^{-1}AP=\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & 3

\end{bmatrix}.\]

Then if we have the equality $B^2=A$, then we need to have

\[(P^{-1}BP)(P^{-1}BP)=p^{-1}AP=\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & 3

\end{bmatrix}.\]
From this we see that the matrix

\[B=P\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & \sqrt{3}

\end{bmatrix}P^{-1}\]
satisfies the above relation.

In fact, we compute

\begin{align*}

B^2&=\left( P\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & \sqrt{3}

\end{bmatrix}P^{-1} \right) \left(P\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & \sqrt{3}

\end{bmatrix}P^{-1}\right) \\[6pt]
&=P\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & \sqrt{3}

\end{bmatrix}^2P^{-1}=P\begin{bmatrix}

0 & 0 & 0 \\

0 &1 &0 \\

0 & 0 & 3

\end{bmatrix}P^{-1}=A.

\end{align*}

Remark that the matrix $B$ is real since $P$ is real. Thus the matrix $B$ satisfies the conditions of the problem.

## Remark.

The explicit form of the matrix $P$ was used to assure that the matrix $B$ is real.

So we did not have to compute the matrix $P$ explicitly if we use the fact that the symmetric matrix is diagonalizable by a real orthogonal matrix.

### Generalization.

See problem A square root matrix of a symmetric matrix with non-negative eigenvalues for a more general question than part (b).

Furthermore, try the next problem.

**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

## 1 Response

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