# 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.

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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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