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.

Consider the $2\times 2$ real matrix
\[A=\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}.\]

(a) Prove that the matrix $A$ is positive definite.

(b) Since $A$ is positive definite by part (a), the formula
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans} A \mathbf{y}\] for $\mathbf{x}, \mathbf{y} \in \R^2$ defines an inner product on $\R^n$.
Consider $\R^2$ as an inner product space with this inner product.

Prove that the unit vectors
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix} \text{ and } \mathbf{e}_2=\begin{bmatrix}
0 \\
1
\end{bmatrix}\] are not orthogonal in the inner product space $\R^2$.

(c) Find an orthogonal basis $\{\mathbf{v}_1, \mathbf{v}_2\}$ of $\R^2$ from the basis $\{\mathbf{e}_1, \mathbf{e}_2\}$ using the Gram-Schmidt orthogonalization process.

Proof.

(a) Prove that the matrix $A$ is positive definite.

We prove that for every nonzero vector $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}\in \R^2$, we have $\mathbf{x}^{\trans} A \mathbf{x} > 0$.
We have
\begin{align*}
\mathbf{x}^{\trans} A \mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix} \begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
x+y \\
x+3y
\end{bmatrix}\\[6pt] &=x(x+y)+y(x+3y)=x^2+2xy+3y^2\\
&=x^2+2xy+y^2+2y^2=(x+y)^2+2y^2.
\end{align*}

Since $\mathbf{x}\neq \mathbf{0}$, at least one of $x, y$ is nonzero.
Thus the last expression is always positive.
Hence $A$ is a positive definite matrix.

(b) Prove that $\mathbf{e}_1, \mathbf{e}_2$ are not orthogonal in the inner product space $\R^2$.

Note that by post “A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space“, the formula $\langle \mathbf{x}, \mathbf{y}\rangle$ defines an inner product on $\R^2$.

Two vectors $\mathbf{x}$ and $\mathbf{y}$ is said to be orthogonal if $\langle \mathbf{x}, \mathbf{y}\rangle=0$.

The vectors $\mathbf{e}_1, \mathbf{e}_2$ are not orthogonal with this inner product since
\begin{align*}
\langle \mathbf{e}_1, \mathbf{e}_2\rangle=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
0 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 \\
3
\end{bmatrix}=1\neq 0.
\end{align*}

(c) Find an orthogonal basis using the Gram-Schmidt orthogonalization process.

By the Gram-Schmidt orthogonalization process, we have
\begin{align*}
\mathbf{v}_1&=\mathbf{e}_1\\
\mathbf{v}_2&=\mathbf{e}_2-\frac{\langle \mathbf{v}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle}\mathbf{v}_1
=\mathbf{e}_2-\frac{\langle \mathbf{e}_1, \mathbf{e}_2 \rangle}{\langle \mathbf{e}_1, \mathbf{e}_1 \rangle}\mathbf{e}_1.
\end{align*}

We compute
\begin{align*}
\langle \mathbf{e}_1, \mathbf{e}_1 \rangle=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
1 \\
0
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 \\
1
\end{bmatrix}=1.
\end{align*}
We also have $\langle \mathbf{e}_1, \mathbf{e}_2\rangle=1$ from part (b).
Thus, we have
\begin{align*}
\mathbf{v}_2=\mathbf{e}_2-\mathbf{e}_1=\begin{bmatrix}
-1 \\
1
\end{bmatrix}.
\end{align*}
Thus, the Gram-Schmidt orthogonalization process yields the orthogonal basis
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
-1 \\
1
\end{bmatrix}.\]

Double Check

Let us verify that $\mathbf{v}_1, \mathbf{v}_2$ are orthogonal by computing their inner product directly as follows.
We have
\begin{align*}
\langle \mathbf{v}_1, \mathbf{v}_2\rangle=\mathbf{v}_1^{\trans} A\mathbf{v}_2=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
1 & 1\\
1& 3
\end{bmatrix}\begin{bmatrix}
-1 \\
1
\end{bmatrix}=\begin{bmatrix}
1 & 0
\end{bmatrix}\begin{bmatrix}
0 \\
2
\end{bmatrix}=0.
\end{align*}

The post The Inner Product on $\R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization first appeared on Problems in Mathematics.

(a) Suppose that $A$ is an $n\times n$ real symmetric positive definite matrix.
Prove that
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.

(b) Let $A$ be an $n\times n$ real matrix. Suppose that
\[\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}\] defines an inner product on the vector space $\R^n$.

Prove that $A$ is symmetric and positive definite.

Definitions.

Inner Product on a Real Vector Space

Let $V$ be a real vector space. An inner product on $V$ is a function that assigns a real number $\langle \mathbf{u}, \mathbf{v}\rangle$ to each pair of vectors $\mathbf{u}$ and $\mathbf{v}$ in $V$ satisfying the following properties.
For any vectors $\mathbf{u}, \mathbf{v}, \mathbf{w}$ and a real number $r\in \R$,

A Positive-Definite Matrix

A real symmetric $n\times n$ matrix $A$ is called positive definite if $\mathbf{x}^{\trans}A\mathbf{x} > 0$ for each nonzero vector $\mathbf{x}\in \R^n$.

Proof.

(a) If $A$ is positive definite, then $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product

First of all, note that we can write
\[\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}A\mathbf{y}=\mathbf{x}\cdot (A\mathbf{y}),\] where the “dot” is the dot product of $\R^n$.
Thus, $\langle \mathbf{x}, \mathbf{y}\rangle$ is a real number.

We verify the three properties of an inner product.
Since the dot product is commutative, we have
\begin{align*}
\langle \mathbf{x}, \mathbf{y}\rangle&=\mathbf{x}\cdot (A\mathbf{y})= (A\mathbf{y})\cdot \mathbf{x}\\
&=(A\mathbf{y})^{\trans}\mathbf{x}=\mathbf{y}^{\trans}A^{\trans}\mathbf{x}\\
&=\mathbf{y}^{\trans}A\mathbf{x} &&\text{since $A$ is symmetric}\\
&=\langle \mathbf{y}, \mathbf{x} \rangle.
\end{align*}
Thus, the function $\langle\,,\,\rangle$ is symmetric.

Next, for any vectors $\mathbf{x}, \mathbf{y}, \mathbf{z}$ and any real number $r$, we have
\begin{align*}
\langle r\mathbf{x}, \mathbf{y}\rangle &=(r\mathbf{x})^{\trans}A\mathbf{y}=r\mathbf{x}^{\trans}A\mathbf{y}=r\langle \mathbf{x}, \mathbf{y}\rangle
\end{align*}
and
\begin{align*}
\langle \mathbf{x}+\mathbf{y}, \mathbf{z}\rangle &=(\mathbf{x}+\mathbf{y})^{\trans}A\mathbf{z}=(\mathbf{x}^{\trans}+\mathbf{y}^{\trans})A\mathbf{z}\\
&=\mathbf{x}^{\trans}A\mathbf{z}+\mathbf{y}^{\trans}A\mathbf{z}=\langle \mathbf{x}, \mathbf{z}\rangle+\langle \mathbf{y}, \mathbf{z}\rangle.
\end{align*}
Thus, the linearity in the first argument is satisfied.

If $\mathbf{x}$ is a nonzero vector in $\R^n$, then we have
\begin{align*}
\langle \mathbf{x}, \mathbf{x}\rangle=\mathbf{x}^{\trans}A\mathbf{x} > 0
\end{align*}
since $A$ is positive definite.
We also have
\begin{align*}
\langle \mathbf{0}, \mathbf{0}\rangle=\mathbf{0}^{\trans}A\mathbf{0}=0.
\end{align*}
It follows that $\langle \mathbf{x}, \mathbf{x}\rangle \geq 0$ for any vector $\mathbf{x}\in \R^n$.

Suppose that $\langle \mathbf{x}, \mathbf{x}\rangle=0$.
Then we have
\begin{align*}
\langle \mathbf{x}, \mathbf{x}\rangle=\mathbf{x}^{\trans}A\mathbf{x}=0.
\end{align*}
Since $A$ is positive definite, this happens if and only if $\mathbf{x}=\mathbf{0}$.
Hence $\langle \mathbf{x}, \mathbf{x}\rangle=0$ if and only if $\mathbf{x}=0$.
This proves the positive-definiteness of the function $\langle\,,\,\rangle$.

This completes the verification of the three properties, and hence $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product on $\R^n$.

(b) If $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ defines an inner product, then $A$ is symmetric positive definite

Suppose that $\langle \mathbf{x}, \mathbf{y}\rangle:=\mathbf{x}^{\trans}A\mathbf{y}$ is an inner product on $\R^n$.
Let us write $A=(a_{ij})$.

We first prove that $A$ is symmetric.
Let $\mathbf{e}_i$ denote the $i$-th standard unit vectors, that is,
\[\mathbf{e}_i=\begin{bmatrix}
0 \\
\vdots \\
1 \\
\vdots \\
0
\end{bmatrix},\] where $1$ is on the $i$-th columns and other entries are all zero.

We compute
\begin{align*}
\langle \mathbf{e}_i, \mathbf{e}_j\rangle &=\mathbf{e}^{\trans}_iA\mathbf{e}_j\\
&=\begin{bmatrix}
0 & \dots &1 & \dots 0
\end{bmatrix}
\begin{bmatrix}
a_{1 j} \\
a_{2 j} \\
\vdots \\
a_{n j}
\end{bmatrix}=a_{ij}.
\end{align*}
Similarly, $\langle \mathbf{e}_j, \mathbf{e}_i\rangle =a_{j i}$. By symmetry of the inner product, it yields that
\[a_{ij}=\langle \mathbf{e}_i, \mathbf{e}_j\rangle =\langle \mathbf{e}_j, \mathbf{e}_i\rangle =a_{j i}.\] Therefore, the matrix $A$ is symmetric.

Next, we show that $A$ is positive definite.
Let $\mathbf{x}$ be a nonzero vector in $\R^n$.
Then we have
\[\mathbf{x}^{\trans}A\mathbf{x}=\langle \mathbf{x}, \mathbf{x}\rangle > 0\] by positive-definiteness property of the inner product.
This proves that $A$ is a positive definite matrix.

Related Question.

A concrete example of a positive-definite matrix is given in the next problem.

See the post ↴
The Inner Product on $\R^2$ induced by a Positive Definite Matrix and Gram-Schmidt Orthogonalization
for proofs.

The post A Symmetric Positive Definite Matrix and An Inner Product on a Vector Space 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.