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Similarity and Diagonalisation

Similar Matrices

The Definition of Similar Matrices

For n-th order square matrices $A$ and $B$ , if there exists an invertible matrix $P$ such that $B = P^{-1}AP$ , then the matrices $A$ and $B$ are called similar, written $A \sim B$ .

Properties of Similar Matrices

Property 1: similarity is an equivalence relation (reflexive, symmetric and transitive).

Property 2: similar matrices have the same characteristic polynomial.

Property 3: similar matrices have the same eigenvalues.

Property 4: similar matrices have the same trace and determinant.

Property 5: similar matrices have the same rank.

The Proof for Similar Matrices

Proof of Property 2: Let $B = P^{-1}AP$ . Then: $|B - \lambda I| = |P^{-1}AP - \lambda I| = |P^{-1}AP - P^{-1}\lambda IP| = |P^{-1}(A - \lambda I)P| = |P^{-1}||A - \lambda I||P| = |A - \lambda I|$

Diagonalisation

The Definition of Diagonalisation

If the matrix $A$ is similar to a diagonal matrix, that is, if there exist an invertible matrix $P$ and a diagonal matrix $D$ such that: $A = PDP^{-1}$

then the matrix $A$ is called diagonalisable.

The Conditions for Diagonalisation

The Necessary and Sufficient Condition for Diagonalisability

An n-th order matrix $A$ is diagonalisable if and only if $A$ has $n$ linearly independent eigenvectors.

Proof

Proof:

Sufficiency: suppose $A$ has $n$ linearly independent eigenvectors $\vec{x}_1, \vec{x}_2, \dots, \vec{x}_n$ with corresponding eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$ . Let $P = [\vec{x}_1, \vec{x}_2, \dots, \vec{x}_n]$ and $D = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n)$ . Then $A = PDP^{-1}$ .
Necessity: if $A = PDP^{-1}$ , then $AP = PD$ , that is, the columns of $A$ are linear combinations of the columns of $P$ , with coefficients given by the diagonal entries of $D$ .

The Steps of Diagonalisation

Steps:

Find the eigenvalues: solve the characteristic equation $\lvert A - \lambda I\rvert = 0$
Find the eigenvectors: for each eigenvalue, find the corresponding linearly independent eigenvectors
Construct the matrices: $P$ is formed from the eigenvectors and $D$ from the eigenvalues
Verify: check that $A = PDP^{-1}$ holds

Example

Example 1: decide whether the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is diagonalisable, and if so find its diagonalisation.

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:
- $\lambda_1 = 1$ : $\vec{x}_1 = (1, -1)$
- $\lambda_2 = 3$ : $\vec{x}_2 = (1, 1)$
Construct the matrices:
- $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
- $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
The diagonalisation: $A = PDP^{-1}$

The Case Where Diagonalisation Fails

The Condition for Failure

The Condition for a Matrix Not to Be Diagonalisable

If the matrix $A$ has a repeated eigenvalue whose geometric multiplicity is smaller than its algebraic multiplicity, then $A$ is not diagonalisable.

Example

Example 2: decide whether the matrix $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is diagonalisable.

Solution:

The characteristic equation: $\lvert A - \lambda I\rvert = (1-\lambda)^2 = 0$
The eigenvalue: $\lambda = 1$ (a double root)
The geometric multiplicity: solving $(A - I)\vec{x} = \vec{0}$ gives $\vec{x} = (1, 0)$ , so the geometric multiplicity is 1
The algebraic multiplicity is 2 and the geometric multiplicity is 1, so the matrix is not diagonalisable

Applications of Diagonalisation

Computing Powers of a Matrix

If $A = PDP^{-1}$ , then $A^n = PD^nP^{-1}$ .

Proof

Proof: $A^n = (PDP^{-1})^n = PDP^{-1} \cdot PDP^{-1} \cdot \dots \cdot PDP^{-1} = PD^nP^{-1}$

Computing Functions of a Matrix

If $A = PDP^{-1}$ , then $f(A) = Pf(D)P^{-1}$ , where $f(D) = \operatorname{diag}(f(\lambda_1), f(\lambda_2), \dots, f(\lambda_n))$ .

The Jordan Canonical Form

The Definition of the Jordan Canonical Form

For a real symmetric matrix $A$ , if there exists an orthogonal matrix $P$ such that $P^TAP = D$ is a diagonal matrix, then $A$ is called diagonalisable, where $D$ is the diagonal matrix whose entries on the diagonal are the eigenvalues of $A$ .

Jordan Blocks

A matrix of the form $J_i = \begin{pmatrix} \lambda_i & 1 & 0 & \dots & 0 \\ 0 & \lambda_i & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & \lambda_i \end{pmatrix}$

is called a Jordan block.

Exercises

Exercise 1

Decide whether the matrix $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is diagonalisable.

Reference Answer

similar matricesdiagonalisationdiagonalisableeigenvector

Approach: Check whether there are enough linearly independent eigenvectors.

Steps:

The characteristic equation: $\lvert A - \lambda I\rvert = (1-\lambda)^2 = 0$
The eigenvalue: $\lambda = 1$ (a double root)
Find the eigenvectors: solve $(A - I)\vec{x} = \vec{0}$
$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
This gives the eigenvector $\vec{x} = (1, 0)$
There is only one linearly independent eigenvector, fewer than the order of the matrix, which is 2

Answer: it is not diagonalisable

Exercise 2

Find the diagonalisation of the matrix $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$ .

Reference Answer

similar matricesdiagonalisationdiagonalisablecharacteristic equation

Approach: Follow the steps of diagonalisation.

Steps:

The characteristic equation: $\lvert A - \lambda I\rvert = (3-\lambda)(2-\lambda) = 0$
The eigenvalues: $\lambda_1 = 2$ , $\lambda_2 = 3$
The eigenvectors:
- $\lambda_1 = 2$ : $\vec{x}_1 = (1, -1)$
- $\lambda_2 = 3$ : $\vec{x}_2 = (1, 0)$
Construct the matrices:
- $P = \begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$
- $D = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$
The diagonalisation: $A = PDP^{-1}$

Answer: $A = PDP^{-1}$ , where $P = \begin{pmatrix} 1 & 1 \\ -1 & 0 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$

Exercise 3

Prove that: similar matrices have the same eigenvalues.

Reference Answer

similar matricesdiagonalisationdiagonalisablecharacteristic polynomial

Approach: Use the property that similar matrices have the same characteristic polynomial.

Steps:

Let $A \sim B$ , that is, $B = P^{-1}AP$
Similar matrices have the same characteristic polynomial: $\lvert A - \lambda I\rvert = \lvert B - \lambda I\rvert$
The eigenvalues are the roots of the characteristic polynomial
Therefore $A$ and $B$ have the same eigenvalues

Answer: the proof is complete

Exercise 4

Let $A = PDP^{-1}$ . Find $A^n$ .

Reference Answer

similar matricesdiagonalisationdiagonalisablematrix

Approach: Use the property of powers of a diagonalised matrix.

Steps:

$A = PDP^{-1}$
$A^n = (PDP^{-1})^n = PD^nP^{-1}$
$D^n = \operatorname{diag}(\lambda_1^n, \lambda_2^n, \dots, \lambda_n^n)$

Answer: $A^n = PD^nP^{-1}$ , where $D^n = \operatorname{diag}(\lambda_1^n, \lambda_2^n, \dots, \lambda_n^n)$

Exercise 5

Decide whether the matrix $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ is diagonalisable, and if so find its diagonalisation.

Reference Answer

similar matricesdiagonalisationdiagonalisableeigenvector

Approach: Follow the steps of diagonalisation.

Steps:

The characteristic equation: $\lvert A - \lambda I\rvert = \lambda^2 - 4\lambda + 3 = 0$
The eigenvalues: $\lambda_1 = 1$ , $\lambda_2 = 3$
The eigenvectors:
- $\lambda_1 = 1$ : $\vec{x}_1 = (1, -1)$
- $\lambda_2 = 3$ : $\vec{x}_2 = (1, 1)$
There are two linearly independent eigenvectors, equal to the order of the matrix
Construct the matrices:
- $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
- $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$
The diagonalisation: $A = PDP^{-1}$

Answer: it is diagonalisable, $A = PDP^{-1}$ , where $P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$

Summary

Symbols Used in This Article

Symbol	Type	Pronunciation / Note	Meaning in This Article
$P^{-1}AP$	Mathematical symbol	P inverse A P	The form of a similarity transformation: $B = P^{-1}AP$
$A \sim B$	Mathematical symbol	A tilde B	The matrices $A$ and $B$ are similar
$P$	Mathematical symbol	invertible matrix P	The invertible matrix formed from the linearly independent eigenvectors
$A = PDP^{-1}$	Mathematical symbol	A equals P D P inverse	The diagonalisation decomposition
$D = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$	Mathematical symbol	diagonal matrix of lambdas	The diagonal matrix whose diagonal entries are the eigenvalues
$\lambda_i$	Greek letter	Lambda sub i	The $i$ -th eigenvalue
$\lvert A - \lambda I\rvert$	Mathematical symbol	determinant of (A minus lambda I)	The characteristic polynomial, which similar matrices share
$A^n = PD^nP^{-1}$	Mathematical symbol	A to the n equals P D to the n P inverse	The computation of powers of a matrix after diagonalisation
$J_i$	Mathematical symbol	Jordan block J i	A Jordan block, an upper triangular matrix with 1s on the superdiagonal

Visualising Eigenvectors

Real Symmetric Matrices

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Similarity and Diagonalisation

Similar Matrices

The Definition of Similar Matrices

Properties of Similar Matrices

The Proof for Similar Matrices

Diagonalisation

The Definition of Diagonalisation

The Conditions for Diagonalisation

The Steps of Diagonalisation

Example

The Case Where Diagonalisation Fails

The Condition for Failure

Example

Applications of Diagonalisation

Computing Powers of a Matrix

Computing Functions of a Matrix

The Jordan Canonical Form

The Definition of the Jordan Canonical Form

Jordan Blocks

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Summary

Symbols Used in This Article

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