For each of the following matrix $A$, prove that $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ for all vectors $\mathbf{x}$ in $\R^2$. Also, determine those vectors $\mathbf{x}\in \R^2$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$.

(a) $A=\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}$.

 
(b) $A=\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}$.

Proof.

(a) $A=\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}$ be a vector in $\R^2$. Then we have
\begin{align*}
\mathbf{x}^{\trans}A\mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
4x+2y \\
2x+y
\end{bmatrix}\\[6pt] &=x(4x+2y)+y(2x+y)=4x^2+2xy+2xy+y^2\\
&=4x^2+4xy+y^2\\
&=(2x+y)^2 \geq 0.
\end{align*}
Thus we have $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$.
Note that $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if $2x+y=0$.
Thus those vectors $\mathbf{x}$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$ are
\[\mathbf{x}=\begin{bmatrix}
x \\
-2x
\end{bmatrix}=x\begin{bmatrix}
1 \\
-2
\end{bmatrix}\] for any real number $x$.

(b) $A=\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}$ be a vector in $\R^2$.
We compute
\begin{align*}
\mathbf{x}^{\trans}A\mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}=
\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
2x+y \\
x+3y
\end{bmatrix}\\[6pt] &=x(2x+y)+y(x+3y)=2x^2+xy+xy+3y^2\\
&=2x^2+2xy+3y^2\\
&=x^2+(x+y)^2+2y^2 \geq 0. \tag{*}
\end{align*}
Thus we obtain $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$.
It follows from (*) that $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if
\[x^2+(x+y)^2+2y^2=0.\] Since each of $x^2, (x+y)^2, 2y^2$ is nonnegative, we must have $x=y=0$.
Therefore $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if $\mathbf{x}=\mathbf{0}$.

The post Prove $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ and determine those $\mathbf{x}$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$ first appeared on Problems in Mathematics.

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.

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).

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.

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.

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.

The post A Positive Definite Matrix Has a Unique Positive Definite Square Root first appeared on Problems in Mathematics.

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

Find all the square roots of the matrix
\[A=\begin{bmatrix}
2 & 2\\
2& 2
\end{bmatrix}.\]

Proof.

Diagonalize $A$.

We first diagonalize the matrix $A$.
The characteristic polynomial of the matrix $A$ is
\begin{align*}
p(t)=\det(A-tI)=\begin{vmatrix}
2-t & 2\\
2& 2-t
\end{vmatrix}=t(t-4).
\end{align*}
Thus, the eigenvalues of $A$ are $0, 4$.
(Since $A$ has two distinct eigenvalues, it is diagonalizable.)

Let us find eigenvectors.
For the eigenvalue $0$, solving $A\mathbf{x}=\mathbf{0}$, we see that
\[\mathbf{v}=\begin{bmatrix}
1 \\
-1
\end{bmatrix}\] is an eigenvector for $0$.

For the eigenvalue $4$, solving $(A-4I)\mathbf{x}=\mathbf{0}$ yields that
\[\mathbf{u}=\begin{bmatrix}
1 \\
1
\end{bmatrix}\] is an eigenvector for $4$.

Thus the matrix
\[S=\begin{bmatrix}
\mathbf{v} & \mathbf{u}
\end{bmatrix}=\begin{bmatrix}
1 & 1\\
-1& 1
\end{bmatrix}\] diagonalizes the matrix $A$, that is,
\[S^{-1}AS=D,\] where $D$ is the diagonal matrix
\[D=\begin{bmatrix}
0 & 0\\
0& 4
\end{bmatrix}.\]

Determine Square Roots of $A$.

Now suppose that $B$ is a matrix such that $B^2=A$.
We have
\begin{align*}
D=S^{-1}AS=S^{-1}B^2S=(S^{-1}BS)(S^{-1}BS)=(S^{-1}BS)^2=B’^2,
\end{align*}
where we set $B’=S^{-1}BS$.

Observe that
\[B’D=B’B’^2=B’^3=B’^2B’=DB’.\] Since $B’$ commutes with the diagonal matrix $D$, the matrix $B’$ is also diagonal.
(To see this directly, put $B’=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}$ and compute $B’D$ and $DB’$. Then $B’D=D’B’$ requires $b=c=0$.)

Let $B’=\begin{bmatrix}
a & 0\\
0& d
\end{bmatrix}$.
Since $B’^2=D$, we have
\[\begin{bmatrix}
a^2 & 0\\
0& d^2
\end{bmatrix}=\begin{bmatrix}
0 & 0\\
0& 4
\end{bmatrix},\] hence $a=0$ and $d=\pm 2$.

It follows that a square root of $A$ must be $B=SB’S^{-1}$, where $B’$ is one of
\[\begin{bmatrix}
0 & 0\\
0& 2
\end{bmatrix}, \quad \begin{bmatrix}
0 & 0\\
0& -2
\end{bmatrix}.\]

When $B’=\begin{bmatrix}
0 & 0\\
0& 2
\end{bmatrix}$, we compute
\begin{align*}
B&=SB’S^{-1}=\begin{bmatrix}
1 & 1\\
-1& 1
\end{bmatrix}\begin{bmatrix}
0 & 0\\
0& 2
\end{bmatrix}
\frac{1}{2}\begin{bmatrix}
1 & -1\\
1& 1
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix}.
\end{align*}

Similarly, when $B’=\begin{bmatrix}
0 & 0\\
0& -2
\end{bmatrix}$, we obtain
\[B=\begin{bmatrix}
-1 & -1\\
-1& -1
\end{bmatrix}.\]

In summary, the square roots of the matrix $A$ are
\[\begin{bmatrix}
1 & 1\\
1& 1
\end{bmatrix} \text{ and } \begin{bmatrix}
-1 & -1\\
-1& -1
\end{bmatrix}.\]

Related Question.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

The post Find All the Square Roots of a Given 2 by 2 Matrix first appeared on Problems in Mathematics.