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

## 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*)

Add to solve later

## 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

Add to solve later

Sponsored Links