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5.C Matrix square roots

In this section we will discuss a way of defining a “square root” of a matrix. Recall that a square root of a number $a\in{\mathbb{C}}$ (or more generally, we could take $a\in F$ any field) is another number $b\in{\mathbb{C}}$ such that $b^{2}=a$ . As you know, if $a\in{\mathbb{R}}$ then its square roots are only real when $a\geq 0$ , and even then they are not unique. Nevertheless we have the following theorem.

Theorem 5.18.

If $a\in{\mathbb{R}}$ is a non-negative number (which means $a\geq 0$ ), then $a$ has a unique non-negative square root.

In this section, we will generalize the above theorem to matrices, where we replace “non-negative number” with “postive semi-definite matrix”. There are several competing ways to generalize the concept of a “square root” to matrices, but in this module we will only focus on the following one.

Definition 5.19:

Given a matrix $A\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ , the matrix square root of $A$ is a matrix $B\in\operatorname{M}_{{n}}({{\mathbb{C}}})$ such that

A=B^{2}.

The following exercise shows that matrix square roots don’t always exist:

Exercise 5.20:

Prove that there is no matrix $B\in\operatorname{M}_{{2}}({{\mathbb{C}}})$ such that $B^{2}=\begin{bmatrix}0&1\\ 0&0\end{bmatrix}$ .

[End of Exercise]

Below we will see the following analogy: Postive real numbers are to positive definite matrices, as non-negative real numbers are to positive semi-definite matrices. A matrix $A$ is positive semi-definite if:

\vec{x}^{T}A\vec{x}\geq 0

for any non-zero vector $0\neq\vec{x}\in{\mathbb{R}}^{n}$ .

So positive definite matrices are also positive semi-definite. This concept occurs naturally in probability and statistics; for example, the covariance matrix of $n$ random variables is always positive semi-definite (see MATH230).

Theorem 5.21.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ be real symmetric. The following are equivalent:

i.

$A$ is positive semi-definite,
ii.

All of the eigenvalues of $A$ are non-negative (i.e. $\geq 0$ ).

In the above theorem Sylvester’s criterion does not appear because it is no longer valid; in other words, being real, symmetric and positive semi-definite is not equivalent to being real, symmetric and having all principal minors $\geq 0$ . The only reliable test is the eigenvalue test.

Proof.

The proof is similar to the proof of Theorem 5.15. ∎

Exercise 5.22:

Verify that the matrix $\begin{bmatrix}1&0&0\\ 0&2&-2\\ 0&-2&2\end{bmatrix}$ is symmetric and positive semi-definite, but not positive definite.

[End of Exercise]

Theorem 5.23.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ be a real symmetric positive semi-definite matrix. Then there exists a unique real symmetric positive semi-definite matrix $B$ such that $A=B^{2}$

In this case, the resulting matrix is usually called “the” matrix square root of $A$ , since it’s uniquely defined. So, in this way, “real symmetric positive semi-definite matrices” may be considered as a nice generalization of “non-negative real numbers”.

Proof.

There is an orthogonal matrix $P$ and diagonal matrix $D$ such that

A=PDP^{T}.

This is the Spectral Theorem 5.7. Since $A$ is positive semi-definite, all of the diagonal entries of $D$ are non-negative (i.e. $\lambda_{i}\geq 0$ ), so we can define $C$ as follows

$D=\operatorname{diag}(\lambda_{1},\cdots,\lambda_{n})$ ,
$C:=\operatorname{diag}(\sqrt{\lambda_{1}},\cdots,\sqrt{\lambda_{n}}).$

Then $C^{2}=D$ , and $B:=PCP^{T}$ is real symmetric positive semi-definite. Finally,

B^{2}=(PCP^{T})(PCP^{T})=PC(P^{T}P)CP^{T}=PC^{2}P^{T}=PDP^{T}=A.

Therefore, we have proved that such a $B$ always exists.

We omit the proof of uniqueness (the proof is not obvious). ∎

Example 5.24.

Find the matrix square root of $A$ from Example 5.9.

In that example we found an orthogonal $P$ and diagonal $D$ such that $A=PDP^{T}$ . By taking the square root of the diagonal entries of $D$ , we compute:

B=P\sqrt{D}P^{T}={\begin{bmatrix}\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{2}}&\frac{-% 1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}&0&\frac{2}{\sqrt{6}}\\ \frac{1}{\sqrt{3}}&\frac{-1}{\sqrt{2}}&\frac{-1}{\sqrt{6}}\end{bmatrix}}\begin% {bmatrix}2&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}{\begin{bmatrix}\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}}&\frac{% 1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}&0&\frac{-1}{\sqrt{2}}\\ \frac{-1}{\sqrt{6}}&\frac{2}{\sqrt{6}}&\frac{-1}{\sqrt{6}}\end{bmatrix}}=\frac% {1}{3}\begin{bmatrix}4&1&1\\ 1&4&1\\ 1&1&4\end{bmatrix}.

Now it is easy to check that $B^{2}=A$ .