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}
\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.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

The post A Square Root Matrix of a Symmetric Matrix first appeared on Problems in Mathematics.