# A Square Root Matrix of a Symmetric Matrix with Non-Negative Eigenvalues ## Problem 63

Let $A$ be an $n\times n$ real symmetric matrix whose eigenvalues are all non-negative real numbers.

Show that there is an $n \times n$ real matrix $B$ such that $B^2=A$. Add to solve later

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## Hint.

Use the fact that a real symmetric matrix is diagonalizable by a real orthogonal matrix.

## Proof.

Since $A$ is a real symmetric matrix, it can be diagonalizable by a real orthogonal matrix $P$.
Thus we have
$P^{-1}AP=D,$ where $D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $A$.

The matrix $D$ is real since all the eigenvalues are real.
(Note that this is always true for a real symmetric matrix.)

If we have $B^2=A$ for some $B$, then we have
$(P^{-1}BP)(P^{-1}BP)=P^{-1}AP=D.$

From this, we see that we can choose $B=PD’P^{-1}$ where $D’$ is the diagonal matrix whose $i$-th diagonal entry is the square root of $i$-th diagonal entries of $D$.
The matrix $D’$ is real since the eigenvalues are non-negative by the assumption, hence the square roots of them are still real.

Since $P$ is also real, the matrix $B$ is real and satisfies
$B^2=(PD’P^{-1})(PD’P^{-1})=PD’^2P^{-1}=PDP^{-1}=A.$

Therefore the matrix $B$ satisfies the conditions of the problem.

## Related Question.

This problems is a generalization of a liner algebra exam problem of Princeton University.

See part (b) of problem A square root matrix of a symmetric matrix.

Another related problem is the following.

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

1. 08/09/2016

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