# A Positive Definite Matrix Has a Unique Positive Definite Square Root

## Problem 514

Prove that a positive definite matrix has a unique positive definite square root.

In this post, we review several definitions (a square root of a matrix, a positive definite matrix) and solve the above problem.

After the proof, several extra problems about square roots of a matrix are given.

## Definitions (Square Roots of a Matrix)

Let $A$ be a square matrix. If there exists a matrix $B$ such that $B^2=A$, then we call $B$ a square root of the matrix $A$.

### Examples

For example, if $A=\begin{bmatrix} 2 & 2\\ 2& 2 \end{bmatrix}$, then it is straightforward to see that
$\begin{bmatrix} 1 & 1\\ 1& 1 \end{bmatrix} \text{ and } \begin{bmatrix} -1 & -1\\ -1& -1 \end{bmatrix}$ are square roots of $A$.
(The less trivial question is that these are the only square roots of $A$.
See the post “Find All the Square Roots of a Given 2 by 2 Matrix” .)

Some matrices do not have a square root at all.
For example, the matrix $A=\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ does not have a square root matrix.
(See the post “No/Infinitely Many Square Roots of 2 by 2 Matrices” Part (a).)

On the other hand, some matrices have infinitely many square roots.
For example, the $2\times 2$ identity matrix has infinitely many distinct square roots.
(See the post “No/Infinitely Many Square Roots of 2 by 2 Matrices” Part (b).)

## Analogy with Positive Real Number

For a positive real number $a$, there are two square roots $\pm \sqrt{a}$.
Here $\sqrt{a}$ is the unique positive number whose square is $a$.

The corresponding notion of positive number in matrices is positive definite.

### Definition (Positive Definite Matrix)

An $n\times n$ real symmetric matrix $A$ is said to be positive definite if $\mathbf{v}^{\trans}A\mathbf{v}$ is positive for all nonzero vector $\mathbf{v}\in \R^n$.

We will use the following fact in the proof.

Fact.
A real symmetric matrix $A$ is positive definite if and only if the eigenvalues of $A$ are all positive.

For a proof of this fact, see the post “Positive definite Real Symmetric Matrix and its Eigenvalues”.

### Problem

Just like a positive real number $a$ has a unique positive square root $\sqrt{a}$, we can prove the following (which is the problem of this post).

Problem.
A positive definite matrix has a unique positive definite square root.

## Proof.

Let $A$ be a positive definite $n\times n$ matrix.

### Existence of a Square Root

We first show that there exists a positive definite matrix $B$ such that $B^2=A$.

Let $\lambda_1, \dots, \lambda_n$ be eigenvalues of $A$.
Since $A$ is a real symmetric matrix, it is diagonalizable by an orthogonal matrix $S$.
That is, we have
$S^{\trans}AS=D,$ where $D$ is the diagonal matrix
$D=\begin{bmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 &\lambda_2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}.$ (Note that $S^{-1}=S^{\trans}$ since $S$ is orthogonal.)

Recall that the eigenvalues of a positive definite matrix are positive real numbers by Fact.
So eigenvalues $\lambda_i$ of $A$ are positive real numbers.
Hence $\sqrt{\lambda_i}$ is a positive real number for $i=1, \dots, n$.

Define the matrix
$B=SD’S^{\trans},$ where $D’$ is the diagonal matrix
$D’=\begin{bmatrix} \sqrt{\lambda_1} & 0 & 0 & 0 \\ 0 &\sqrt{\lambda_2} & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sqrt{\lambda_n} \end{bmatrix}.$

Then $B$ is a symmetric matrix because
$B^{\trans}=(SD’S^{\trans})^{\trans}=(S^{\trans})^{\trans}D’^{\trans}S^{\trans}=SD’S^{\trans}=B.$

It follows from the definition of $B=SD’S^{\trans}$ that the eigenvalues of $B$ are positive numbers $\sqrt{\lambda_1}, \dots, \sqrt{\lambda_n}$.
Thus by Fact the matrix $B$ is positive-definite.

The matrix $B$ is a square root of $A$ since we have
\begin{align*}
B^2=(SD’S^{\trans})(SD’S^{\trans})=SD’^2S^{\trans}=SDS^{\trans}=A.
\end{align*}

### Uniqueness of a Square Root

Now we prove the uniqueness of the square root.
Suppose that $C$ is another positive definite square roots of the matrix $A$.

Since $C$ is a real symmetric matrix, there is an orthogonal matrix $P$ such that
$P^{\trans}CP=T.$ Here $T$ is the diagonal matrix
$T=\begin{bmatrix} \mu_1 & 0 & 0 & 0 \\ 0 &\mu_2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mu_n \end{bmatrix},$ where $\mu_1, \dots, \mu_n$ are eigenvalues of $C$.

Since $C^2=A$, we have
\begin{align*}
P^{\trans}AP&=P^{\trans}C^2P=(P^{\trans}CP)^2=T^2\6pt] &=\begin{bmatrix} \mu_1^2& 0 & 0 & 0 \\ 0 &\mu_2^2 & 0 & 0 \\ \vdots & \cdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mu_n^2 \end{bmatrix}. \end{align*} Thus, the matrix P diagonalizes A, and it follows that up to permutation \mu_1^2, \dots, \mu_n^2 are equal to \lambda_1, \dots, \lambda_n. Hence we can modify P (by interchanging columns vectors), and without loss of generality we may assume that \mu_i^2=\lambda_i for i=1, \dots, n. Thus P^{\trans}CP=D’ or equivalently, \[ C=PD’P^{\trans}.

Since $B^2=A=C^2$, we have
\begin{align*}
&(SD’S^{\trans})^2=(PD’P^{\trans})^2\\
&\Leftrightarrow SD’^2S^{\trans}=PD’^2P^{\trans}\\
&\Leftrightarrow SDS^{\trans}=PDP^{\trans}\\
&\Leftrightarrow (P^{\trans}S) D=D (P^{\trans}S).
\end{align*}

Let $Q:=P^\trans S$. Then the last equality is $QD=D Q$, and hence $Q$ commutes with $D$.

Without loss of generality, we may assume that the matrix $D$ can be expressed as a block matrix
$D= \left[\begin{array}{c|c|c|c} \lambda_1 I_1 & 0 &\dots &0\\ \hline 0 & \lambda_2 I_2 & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k I_k \end{array} \right] ,$ where $\lambda_1, \dots, \lambda_k$ are distinct eigenvalues of $A$ and $I_k$ are some identity matrix. (These $\lambda_i$ and the previous $\lambda_i$ are different. The size of the identity matrix $I_j$ is the algebraic multiplicity of $\lambda_j$.)

Express the matrix $Q$ as the block matrix with the same partition as $D$, and write it as
$Q=\left[\begin{array}{c|c|c|c} Q_{11} & Q_{12} &\dots &Q_{1 k}\\ \hline Q_{21} & Q_{22}& \dots & Q_{2k}\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline Q_{k1} & \dots & \dots & Q_{k k} \end{array} \right].$

Comparing the $(i,j)$-block of both sides of $QD=DQ$ yields that
$\lambda_j Q_{ij}=\lambda_i Q_{ij}.$

It follows that if $i\neq j$ then $Q_{ij}$ is the zero matrix.
Thus, $Q$ is a block diagonal matrix
$Q=\left[\begin{array}{c|c|c|c} Q_{11} & 0 &\dots &0\\ \hline 0 & Q_{22}& \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & Q_{k k} \end{array} \right].$ Note that $D’$ is also a block diagonal matrix with the same partition.
It follows that
\begin{align*}
QD’&=\left[\begin{array}{c|c|c|c}
Q_{11} & 0 &\dots &0\\
\hline
0 & Q_{22}& \dots & 0\\
\hline
\vdots & \dots & \ddots& \vdots\\
\hline
0 & \dots & \dots & Q_{k k}
\end{array}
\right] \left[\begin{array}{c|c|c|c}
\lambda_1 I_1 & 0 &\dots &0\\
\hline
0 & \lambda_2 I_2 & \dots & 0\\
\hline
\vdots & \dots & \ddots& \vdots\\
\hline
0 & \dots & \dots & \lambda_k I_k
\end{array}
\right]\6pt] &=\left[\begin{array}{c|c|c|c} \lambda_1 Q_{11} & 0 &\dots &0\\ \hline 0 & \lambda_2 Q_{22} & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k Q_{kk} \end{array} \right]\\[6pt] &=\left[\begin{array}{c|c|c|c} \lambda_1 I_1 & 0 &\dots &0\\ \hline 0 & \lambda_2 I_2 & \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & \lambda_k I_k \end{array} \right] \left[\begin{array}{c|c|c|c} Q_{11} & 0 &\dots &0\\ \hline 0 & Q_{22}& \dots & 0\\ \hline \vdots & \dots & \ddots& \vdots\\ \hline 0 & \dots & \dots & Q_{k k} \end{array} \right] =D’Q. \end{align*} It yields that \[D’=Q^{\trans}D’Q since $Q=P^{\trans}S$ is an orthogonal matrix.

Then we obtain
\begin{align*}
B&=SD’S^{\trans}=S(Q^{\trans}D’Q)S^{\trans}\\
&=SS^{\trans} PD’ P^{\trans} S S^{\trans}\\
&=PD’P^{\trans}=C.
\end{align*}

Therefore, any square root of $A$ must be equal to the matrix $B$.
This completes the proof of the uniqueness, hence the proof of the problem.

## Remark.

The above problem is still true if “positive definite” is replaced by “positive semi-definite”.

## Related Questions About Square Roots of a Matrix

If you want to solve more problems about square roots of a matrix, then try the following problems.

Problem. Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ where
$A=\begin{bmatrix} 1 & -1 & 0 \\ -1 &2 &-1 \\ 0 & -1 & 1 \end{bmatrix}\,\,\,\,?$

This is a part of Princeton University’s Linear Algebra exam problems.
See the post ↴
A Square Root Matrix of a Symmetric Matrix
for a solution.

Problem. 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?

This is one of Berkeley’s qualifying exam problems.
See the post ↴
Square Root of an Upper Triangular Matrix. How Many Square Roots Exist?
for a solution.

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##### Find All the Square Roots of a Given 2 by 2 Matrix

Let $A$ be a square matrix. A matrix $B$ satisfying $B^2=A$ is call a square root of $A$. Find all...

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