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Real Symmetric Matrices

The Definition of a Real Symmetric Matrix

The Definition of a Symmetric Matrix

A matrix $A$ satisfying $A^T = A$ is called a symmetric matrix.

The Definition of a Real Symmetric Matrix

If the matrix $A$ is a real matrix and satisfies $A^T = A$ , then $A$ is called a real symmetric matrix.

Example

Example 1: the matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ is a real symmetric matrix, because $A^T = A$ .

Properties of Real Symmetric Matrices

Basic Properties

Property 1: the eigenvalues of a real symmetric matrix are all real.

Property 2: the eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are orthogonal.

Property 3: a real symmetric matrix is diagonalisable.

Property 4: a real symmetric matrix can be orthogonally diagonalised.

The Proof of Property 1

Proof: let $A$ be a real symmetric matrix, $\lambda$ one of its eigenvalues, and $\vec{x}$ the corresponding eigenvector.

$A\vec{x} = \lambda\vec{x}$

Take the conjugate transpose of both sides: $\vec{x}^*A^T = \lambda^*\vec{x}^*$

Since $A$ is real symmetric, $A^T = A$ , so: $\vec{x}^*A = \lambda^*\vec{x}^*$

Right-multiply both sides by $\vec{x}$ : $\vec{x}^*A\vec{x} = \lambda^*\vec{x}^*\vec{x}$

On the left: $\vec{x}^*A\vec{x} = \vec{x}^*\lambda\vec{x} = \lambda\vec{x}^*\vec{x}$

Therefore: $\lambda\vec{x}^*\vec{x} = \lambda^*\vec{x}^*\vec{x}$

Since $\vec{x} \neq \vec{0}$ , we have $\vec{x}^*\vec{x} > 0$ , so $\lambda = \lambda^*$ , that is, $\lambda$ is real.

The Proof of Property 2

Proof: let $\lambda_1 \neq \lambda_2$ be two distinct eigenvalues of the real symmetric matrix $A$ , and let $\vec{x}_1, \vec{x}_2$ be the corresponding eigenvectors.

$A\vec{x}_1 = \lambda_1\vec{x}_1$ $A\vec{x}_2 = \lambda_2\vec{x}_2$

Left-multiply the first equality by $\vec{x}_2^T$ : $\vec{x}_2^TA\vec{x}_1 = \lambda_1\vec{x}_2^T\vec{x}_1$

Since $A$ is symmetric: $\vec{x}_2^TA\vec{x}_1 = \vec{x}_2^TA^T\vec{x}_1 = (A\vec{x}_2)^T\vec{x}_1 = \lambda_2\vec{x}_2^T\vec{x}_1$

Therefore: $\lambda_1\vec{x}_2^T\vec{x}_1 = \lambda_2\vec{x}_2^T\vec{x}_1$

Since $\lambda_1 \neq \lambda_2$ , we get $\vec{x}_2^T\vec{x}_1 = 0$ , that is, $\vec{x}_1$ and $\vec{x}_2$ are orthogonal.

Orthogonal Diagonalisation

The Definition of Orthogonal Diagonalisation

For a symmetric matrix $A$ , if there exists an orthogonal matrix $P$ such that $P^TAP = D$ is a diagonal matrix, then $A$ is called orthogonally diagonalisable.

The Conditions for Orthogonal Diagonalisation

Real Symmetric Matrices Can Be Orthogonally Diagonalised

A real symmetric matrix can be orthogonally diagonalised.

Proof

Proof:

A real symmetric matrix is diagonalisable
The eigenvectors of a real symmetric matrix can be orthogonalised
Therefore it can be orthogonally diagonalised

The Steps of Orthogonal Diagonalisation

Steps:

Find the eigenvalues: solve the characteristic equation $\lvert A - \lambda I\rvert = 0$
Find the eigenvectors: for each eigenvalue, find the corresponding eigenvectors
Orthogonalise: apply Gram-Schmidt orthogonalisation to the eigenvectors belonging to a repeated eigenvalue
Normalise: normalise all the eigenvectors
Construct the matrices: $Q$ is formed from the normalised eigenvectors and $D$ from the eigenvalues

Example

Example 2: find the orthogonal diagonalisation of the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ .

Solution:

The characteristic equation: $\lvert A - \lambda I\rvert = \lambda^2 - 4\lambda + 3 = 0$
The eigenvalues: $\lambda_1 = 1$ , $\lambda_2 = 3$
The eigenvectors:
- $\lambda_1 = 1$ : $\vec{x}_1 = (1, -1)$
- $\lambda_2 = 3$ : $\vec{x}_2 = (1, 1)$
Normalisation:
- $\vec{q}_1 = \frac{1}{\sqrt{2}}(1, -1)$
- $\vec{q}_2 = \frac{1}{\sqrt{2}}(1, 1)$
Construct the matrices:
- $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
- $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
The orthogonal diagonalisation: $A = QDQ^T$

The Spectral Theorem

The Statement of the Spectral Theorem

The Spectral Theorem

Let $A$ be an n-th order real symmetric matrix. Then there exist an orthogonal matrix $Q$ and a diagonal matrix $D$ such that: $A = QDQ^T$

where the diagonal entries of $D$ are the eigenvalues of $A$ , and the columns of $Q$ are the corresponding unit orthogonal eigenvectors of $A$ .

Applications of the Spectral Theorem

Application 1: powers of a real symmetric matrix $A^n = QD^nQ^T$

Application 2: functions of a real symmetric matrix $f(A) = Qf(D)Q^T$

where $f(D) = \operatorname{diag}(f(\lambda_1), f(\lambda_2), \dots, f(\lambda_n))$ .

Positive Definite Matrices

The Definition of a Positive Definite Matrix

A symmetric matrix $A$ satisfying $A^T = A$ is called a positive definite matrix if $\vec{x}^TA\vec{x} > 0$ for every nonzero vector $\vec{x}$ .

Equivalent Conditions for a Positive Definite Matrix

Let $A$ be an n-th order real symmetric matrix. The following conditions are equivalent:

$A$ is a positive definite matrix
All the eigenvalues of $A$ are positive
All the leading principal minors of $A$ are positive
There exists an invertible matrix $B$ such that $A = B^TB$

Example

Example 3: decide whether the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is positive definite.

Solution:

The eigenvalues: $\lambda_1 = 1$ , $\lambda_2 = 3$ , both positive
The leading principal minors: $\lvert A_1\rvert = 2 > 0$ , $\lvert A_2\rvert = 3 > 0$
Therefore $A$ is a positive definite matrix

Exercises

Exercise 1

Decide whether the matrix $A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ is a real symmetric matrix, and state whether all of its eigenvalues are real.

Reference Answer

symmetric matrixorthogonal diagonalisationreal symmetric matrixeigenvalue

Approach: Check the symmetry of the matrix and its eigenvalues.

Steps:

Check symmetry: $A^T = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = A$ , so it is a real symmetric matrix
Find the eigenvalues: $\lvert A - \lambda I\rvert = \lambda^2 - 1 = 0$
The eigenvalues: $\lambda = 1$ or $\lambda = -1$ , both real

Answer: it is a real symmetric matrix, and all its eigenvalues are real

Exercise 2

Find the orthogonal diagonalisation of the matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ .

Reference Answer

symmetric matrixorthogonal diagonalisationreal symmetric matrixorthogonal

Approach: Follow the steps of orthogonal diagonalisation.

Steps:

The characteristic equation: $\lvert A - \lambda I\rvert = (3-\lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = 0$
The eigenvalues: $\lambda_1 = 2$ , $\lambda_2 = 4$
The eigenvectors:
- $\lambda_1 = 2$ : $\vec{x}_1 = (1, -1)$
- $\lambda_2 = 4$ : $\vec{x}_2 = (1, 1)$
Normalisation:
- $\vec{q}_1 = \frac{1}{\sqrt{2}}(1, -1)$
- $\vec{q}_2 = \frac{1}{\sqrt{2}}(1, 1)$
Construct the matrices:
- $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
- $D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$
The orthogonal diagonalisation: $A = QDQ^T$

Answer: $A = QDQ^T$ , where $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$

Exercise 3

Prove that: the eigenvectors of a real symmetric matrix corresponding to distinct eigenvalues are orthogonal.

Reference Answer

symmetric matrixorthogonal diagonalisationreal symmetric matrixmatrix

Approach: Use the properties of real symmetric matrices.

Steps:

Let $\lambda_1 \neq \lambda_2$ be two distinct eigenvalues of the real symmetric matrix $A$
Let $\vec{x}_1, \vec{x}_2$ be the corresponding eigenvectors
$A\vec{x}_1 = \lambda_1\vec{x}_1$ , $A\vec{x}_2 = \lambda_2\vec{x}_2$
Left-multiply the first equality by $\vec{x}_2^T$ : $\vec{x}_2^TA\vec{x}_1 = \lambda_1\vec{x}_2^T\vec{x}_1$
Since $A$ is symmetric: $\vec{x}_2^TA\vec{x}_1 = \vec{x}_2^TA^T\vec{x}_1 = (A\vec{x}_2)^T\vec{x}_1 = \lambda_2\vec{x}_2^T\vec{x}_1$
Therefore: $\lambda_1\vec{x}_2^T\vec{x}_1 = \lambda_2\vec{x}_2^T\vec{x}_1$
Since $\lambda_1 \neq \lambda_2$ , we get $\vec{x}_2^T\vec{x}_1 = 0$ , that is, $\vec{x}_1$ and $\vec{x}_2$ are orthogonal

Answer: the proof is complete

Exercise 4

Decide whether the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is positive definite.

Reference Answer

symmetric matrixorthogonal diagonalisationreal symmetric matrixeigenvalue

Approach: Check the eigenvalues or the leading principal minors.

Steps:

Method 1: find the eigenvalues
- The characteristic equation: $\lvert A - \lambda I\rvert = \lambda^2 - 4\lambda + 3 = 0$
- The eigenvalues: $\lambda_1 = 1$ , $\lambda_2 = 3$ , both positive
Method 2: check the leading principal minors
- $\lvert A_1\rvert = 2 > 0$
- $\lvert A_2\rvert = 3 > 0$

Answer: it is a positive definite matrix

Exercise 5

Let $A$ be a real symmetric matrix. Prove that: all the eigenvalues of $A$ are real.

Reference Answer

symmetric matrixorthogonal diagonalisationreal symmetric matrixmatrix

Approach: Use the properties of real symmetric matrices.

Steps:

Let $\lambda$ be an eigenvalue of $A$ and $\vec{x}$ the corresponding eigenvector
$A\vec{x} = \lambda\vec{x}$
Take the conjugate transpose of both sides: $\vec{x}^*A^T = \lambda^*\vec{x}^*$
Since $A$ is real symmetric, $A^T = A$ , so: $\vec{x}^*A = \lambda^*\vec{x}^*$
Right-multiply both sides by $\vec{x}$ : $\vec{x}^*A\vec{x} = \lambda^*\vec{x}^*\vec{x}$
On the left: $\vec{x}^*A\vec{x} = \vec{x}^*\lambda\vec{x} = \lambda\vec{x}^*\vec{x}$
Therefore: $\lambda\vec{x}^*\vec{x} = \lambda^*\vec{x}^*\vec{x}$
Since $\vec{x} \neq \vec{0}$ , we have $\vec{x}^*\vec{x} > 0$ , so $\lambda = \lambda^*$ , that is, $\lambda$ is real

Answer: the proof is complete

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$A^T = A$	Mathematical symbol	A transpose equals A	The definition of a symmetric matrix
$\lambda$	Greek letter	Lambda	An eigenvalue of a real symmetric matrix, necessarily real
$\vec{x}$	Mathematical symbol	x vector	An eigenvector
$\vec{x}^*$	Mathematical symbol	x conjugate transpose	The conjugate transpose of the vector $\vec{x}$ , used to prove that the eigenvalues are real
$\vec{x}_2^T\vec{x}_1 = 0$	Mathematical symbol	x2 transpose x1 equals zero	The eigenvectors corresponding to distinct eigenvalues are orthogonal
$A = QDQ^T$	Mathematical symbol	A equals Q D Q transpose	The spectral theorem: the orthogonal diagonalisation decomposition
$\vec{q}_1 = \frac{1}{\sqrt{2}}(1,-1)$	Mathematical symbol	q1 equals one over root two times (1, minus 1)	A normalised eigenvector
$\vec{x}^TA\vec{x} > 0$	Mathematical symbol	x transpose A x greater than zero	The criterion for a positive definite matrix
$A = B^TB$	Mathematical symbol	A equals B transpose B	An equivalent decomposition of a positive definite matrix

Similarity and Diagonalisation

Comprehensive Exercises on Eigenvalues and Eigenvectors

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Real Symmetric Matrices

The Definition of a Real Symmetric Matrix

The Definition of a Symmetric Matrix

The Definition of a Real Symmetric Matrix

Example

Properties of Real Symmetric Matrices

Basic Properties

The Proof of Property 1

The Proof of Property 2

Orthogonal Diagonalisation

The Definition of Orthogonal Diagonalisation

The Conditions for Orthogonal Diagonalisation

The Steps of Orthogonal Diagonalisation

Example

The Spectral Theorem

The Statement of the Spectral Theorem

Applications of the Spectral Theorem

Positive Definite Matrices

The Definition of a Positive Definite Matrix

Equivalent Conditions for a Positive Definite Matrix

Example

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Summary

Symbols Used in This Article

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