Square Root of an Upper Triangular Matrix. How Many Square Roots Exist?

Linear Algebra exam problems and solutions at University of California, Berkeley

Problem 133

Find a square root of the matrix
\[A=\begin{bmatrix}
1 & 3 & -3 \\
0 &4 &5 \\
0 & 0 & 9
\end{bmatrix}.\]

How many square roots does this matrix have?

(University of California, Berkeley Qualifying Exam)
 
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Proof.

We will find all matrices $B$ such that $B^2=A$. Such matrices $B$ are square roots of the matrix $A$.

Note that since $A$ is a diagonal matrix, the eigenvalues of $A$ are diagonal entries $1, 4, 9$. Since $A$ has three distinct eigenvalues, it is diagonalizable.
Solving $(A-\lambda I)\mathbf{x}=\mathbf{0}$ for $\lambda=1,4,9$, we find eigenvectors corresponding to eigenvalues $1, 4, 9$ are respectively

\[ \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \quad
\begin{bmatrix}
1 \\
1 \\
0
\end{bmatrix} , \quad
\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}.\]

Thus the invertible matrix
\[P=\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\] diagonalizes the matrix $A$, that is, we have
\[P^{-1} AP=\begin{bmatrix}
1 & 0 & 0 \\
0 &4 &0 \\
0 & 0 & 9
\end{bmatrix}.\]


Then if $B^2=A$, then we have $(P^{-1}BP)(P^{-1}B)=P^{-1}AP$.
Let $A’=P^{-1}AP$ and $B’=P^{-1}BP$.

Since we have $B’^2=A’$, we have $B’A’=B’^3=A’B’$.
Since $A’$ is diagonal with distinct diagonal entries, this implies that $B’$ is also a diagonal matrix.

A diagonal matrix $B’$ satisfying $B’^2=A’=\begin{bmatrix}
1 & 0 & 0 \\
0 &4 &0 \\
0 & 0 & 9
\end{bmatrix}$ is one of
\[\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}.\] Hence $B$ must be one of
\[P\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}P^{-1}.\] The inverse matrix of $P$ can be calculated as
\[P^{-1}=\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}.\] Therefore, all the square roots of the matrix $A$ are
\[\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
\pm 1 & 0 & 0 \\
0 &\pm 2 &0 \\
0 & 0 & \pm 3
\end{bmatrix}\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}\] and we have $8$ square root matrices.


For example, when the diagonal matrix has all positive entries, then one of the square roots is
\[\begin{bmatrix}
1 & 1 & 0 \\
0 &1 &1 \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 &0 \\
0 & 0 & 3
\end{bmatrix}\begin{bmatrix}
1 & -1 & 1 \\
0 &1 &-1 \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & 1 & -1 \\
0 &2 &1 \\
0 & 0 & 3
\end{bmatrix}.\]

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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