# 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.

1. (a) Consider the eigenvalues and the Jordan canonical form of $A$
2. (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}

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