8.2 Diagonalization and similarity

One eigenvector is useful. A whole basis of eigenvectors is transformative.

If a square matrix has enough linearly independent eigenvectors, then there is a coordinate system in which the matrix becomes diagonal. In that coordinate system, powers, inverses, and many structural questions become almost trivial.

Why diagonalization matters

Suppose a matrix acts on $\mathbb{R}^n$ . In the standard basis, the action may look complicated because different coordinates mix together. But if you build a new basis from eigenvectors, then the action along each basis vector becomes pure scalar multiplication.

That is exactly what a diagonal matrix does.

Definition

Similarity

Two $n\times n$ matrices $A$ and $B$ are similar if there exists an nonsingular matrix $S$ such that

S^{-1}AS=B.

Similarity means that $A$ and $B$ represent the same linear transformation in two different bases.

Definition

Diagonalization and diagonalizability

Let $A$ be an $n\times n$ matrix.

If there exists an invertible matrix $S$ and scalars $\lambda_1,\dots,\lambda_n$ such that

S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n),

then we say that $A$ is diagonalizable and that the displayed equality is a diagonalization of $A$ .

So diagonalization is a special case of similarity in which the target matrix is diagonal.

Eigenvectors are exactly what fill the diagonalization matrix

Theorem

Characterization of diagonalization

Let $A$ be an $n\times n$ matrix, and let

S=[v_1\ v_2\ \cdots\ v_n]

be an invertible matrix built from column vectors $v_1,\dots,v_n$ .

Then the following are equivalent:

each $v_j$ is an eigenvector of $A$ with eigenvalue $\lambda_j$ ;

S^{-1}AS=\operatorname{diag}(\lambda_1,\dots,\lambda_n).

This theorem is the heart of diagonalization. The columns of the change-of-basis matrix are not arbitrary. They must be eigenvectors.

Equivalently, if $D=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ , then

AS=SD.

That equation says:

the first column of AS is $Av_1$ , while the first column of SD is $\lambda_1v_1$ ;
the second column of AS is $Av_2$ , while the second column of SD is $\lambda_2v_2$ ;
and so on.

So the single matrix identity $AS=SD$ packages all eigenvector equations at once.

Theorem

When is a matrix diagonalizable?

An $n\times n$ matrix $A$ is diagonalizable if and only if it has n linearly independent eigenvectors.

This criterion is the one you should remember. Diagonalization is not about guessing a lucky matrix $S$ ; it is about finding a full basis of eigenvectors.

First diagonalization examples

Worked example

A diagonalizable upper-triangular matrix

Let

A= \begin{bmatrix} 1&1&1\\ 0&2&2\\ 0&0&3 \end{bmatrix}.

Suppose we have eigenvectors

u_1= \begin{bmatrix} 1\\0\\0 \end{bmatrix}, \qquad u_2= \begin{bmatrix} 1\\1\\0 \end{bmatrix}, \qquad u_3= \begin{bmatrix} 3\\4\\2 \end{bmatrix},

with eigenvalues 1, 2, and 3, respectively.

If we set

U=[u_1\ u_2\ u_3],

then the three vectors are linearly independent, so $U$ is invertible. Hence

U^{-1}AU=\operatorname{diag}(1,2,3).

The original matrix is not diagonal, but it becomes diagonal in the eigenvector basis.

Worked example

A matrix that is not diagonalizable

Consider

J= \begin{bmatrix} 1&4\\ 0&1 \end{bmatrix}.

Its only eigenvalue is 1, because

\det(J-\lambda I)= \begin{vmatrix} 1-\lambda&4\\ 0&1-\lambda \end{vmatrix} =(1-\lambda)^2.

Now solve $(J-I)x=0$ :

J-I= \begin{bmatrix} 0&4\\ 0&0 \end{bmatrix}.

So $x_2=0$ and $x_1$ is free. The eigenspace is therefore

\operatorname{span}\left\{ \begin{bmatrix} 1\\0 \end{bmatrix} \right\},

which is only one-dimensional. A $2\times2$ matrix needs two linearly independent eigenvectors to be diagonalizable, so $J$ is not diagonalizable.

Similar matrices preserve eigenvalue data

Similarity is not arbitrary conjugation. It preserves the essential eigenvalue structure of a matrix.

Theorem

Similar matrices have the same characteristic polynomial

If $A$ and $B$ are similar, then

p_A(x)=p_B(x).

In particular, $A$ and $B$ have the same eigenvalues.

This is what makes diagonalization meaningful. The diagonal matrix obtained from $A$ is not a different spectral object. It is the same linear transformation written in a basis that exposes its eigenvalues visibly on the diagonal.

Common mistake

Having the same characteristic polynomial is not enough for similarity

Similarity implies equal characteristic polynomials, but the converse is false. Two matrices can share the same characteristic polynomial and still fail to be similar.

For example,

\begin{bmatrix} 0&1\\ 0&0 \end{bmatrix} \qquad\text{and}\qquad \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}

both have characteristic polynomial $x^2$ , but the first matrix is not similar to the zero matrix.

Diagonalization makes powers and inverses easy

Suppose

A=SDS^{-1}, \qquad D=\operatorname{diag}(\lambda_1,\dots,\lambda_n).

Then matrix algebra becomes much simpler.

Theorem

Powers, inverse, and transpose of a diagonalizable matrix

If $A=SDS^{-1}$ with $D$ diagonal, then for each positive integer m,

A^m=SD^mS^{-1}.

If $A$ is invertible, then every diagonal entry of $D$ is nonzero and

A^{-1}=SD^{-1}S^{-1}.

Also, $A^T$ is diagonalizable and has the same eigenvalues as $A$ .

The key point is that diagonal matrices are easy to power:

\operatorname{diag}(\lambda_1,\dots,\lambda_n)^m =\operatorname{diag}(\lambda_1^m,\dots,\lambda_n^m).

Worked example

Compute a power through diagonalization

Let

A= \begin{bmatrix} 2&1&1\\ 1&2&1\\ 1&1&2 \end{bmatrix}.

Suppose $A$ is diagonalized as

A=S\operatorname{diag}(4,1,1)S^{-1}.

Then

A^m=S\operatorname{diag}(4^m,1,1)S^{-1}.

So the hard part is finding the diagonalization once. After that, every positive power is controlled by replacing 4 with $4^m$ and leaving the repeated 1 entries alone.

Quick check

What must the columns of a diagonalizing matrix S be?

Think about the equation $AS=SD$ .

Solution

Answer

Quick check

Can a 3×3 matrix be diagonalizable with only two linearly independent eigenvectors?

Use the characterization theorem.

Solution

Answer

Quick check

If A is similar to D and D is diagonal, do A and D have the same eigenvalues?

Use the similarity theorem.

Solution

Answer

Exercises

Quick check

Suppose A has eigenvectors $v_1,v_2,v_3$ with eigenvalues $2,5,-1$ , and these vectors are linearly independent. What diagonal matrix appears in a diagonalization of A?

Order the diagonal entries to match the chosen order of the eigenvectors.

Solution

Guided solution

Quick check

Why is $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ not diagonalizable?

Check how many linearly independent eigenvectors it has.

Solution

Guided solution

Quick check

If $A=SDS^{-1}$ with $D=\operatorname{diag}(2,-1)$ , what is $A^3$ ?

Use the power rule for diagonalizable matrices.

Solution

Guided solution

Read 8.1 Eigenvalues, eigenvectors, and eigenspaces first if the homogeneous-system viewpoint is not yet solid.

Continue with 8.3 Characteristic polynomials and diagonalization tests for the polynomial tools that decide whether enough eigenvectors exist.

The basis language here also depends on 6.5 Basis and dimension.

8.2 Diagonalization and similarity

MATH1030: Linear algebra I

Why diagonalization matters

Similarity

Diagonalization and diagonalizability

Eigenvectors are exactly what fill the diagonalization matrix

Characterization of diagonalization

When is a matrix diagonalizable?

First diagonalization examples

A diagonalizable upper-triangular matrix

A matrix that is not diagonalizable

Similar matrices preserve eigenvalue data

Similar matrices have the same characteristic polynomial

Having the same characteristic polynomial is not enough for similarity

Diagonalization makes powers and inverses easy

Powers, inverse, and transpose of a diagonalizable matrix

Compute a power through diagonalization

Quick check

What must the columns of a diagonalizing matrix S be?

Answer

Can a 3×3 matrix be diagonalizable with only two linearly independent eigenvectors?

Answer

If A is similar to D and D is diagonal, do A and D have the same eigenvalues?

Answer

Exercises

Suppose A has eigenvectors $v_1,v_2,v_3$ with eigenvalues $2,5,-1$ , and these vectors are linearly independent. What diagonal matrix appears in a diagonalization of A?

Guided solution

Why is $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ not diagonalizable?

Guided solution

If $A=SDS^{-1}$ with $D=\operatorname{diag}(2,-1)$ , what is $A^3$ ?

Guided solution

Section mastery checkpoint

What is the correct criterion for an n×n matrix A to be diagonalizable?

Prerequisites

Key terms in this unit

More notes in this series

8.2 Diagonalization and similarity

MATH1030: Linear algebra I

Why diagonalization matters

Similarity

Diagonalization and diagonalizability

Eigenvectors are exactly what fill the diagonalization matrix

Characterization of diagonalization

When is a matrix diagonalizable?

First diagonalization examples

A diagonalizable upper-triangular matrix

A matrix that is not diagonalizable

Similar matrices preserve eigenvalue data

Similar matrices have the same characteristic polynomial

Having the same characteristic polynomial is not enough for similarity

Diagonalization makes powers and inverses easy

Powers, inverse, and transpose of a diagonalizable matrix

Compute a power through diagonalization

Quick check

What must the columns of a diagonalizing matrix S be?

Answer

Can a 3×3 matrix be diagonalizable with only two linearly independent eigenvectors?

Answer

If A is similar to D and D is diagonal, do A and D have the same eigenvalues?

Answer

Exercises

Suppose A has eigenvectors v1,v2,v3v_1,v_2,v_3v1​,v2​,v3​ with eigenvalues 2,5,−12,5,-12,5,−1, and these vectors are linearly independent. What diagonal matrix appears in a diagonalization of A?

Guided solution

Why is [1101]\begin{bmatrix}1&1\\0&1\end{bmatrix}[10​11​] not diagonalizable?

Guided solution

If A=SDS−1A=SDS^{-1}A=SDS−1 with D=diag⁡(2,−1)D=\operatorname{diag}(2,-1)D=diag(2,−1), what is A3A^3A3?

Guided solution

Related notes

Section mastery checkpoint

Prerequisites

Key terms in this unit

More notes in this series

Suppose A has eigenvectors $v_1,v_2,v_3$ with eigenvalues $2,5,-1$ , and these vectors are linearly independent. What diagonal matrix appears in a diagonalization of A?

Why is $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ not diagonalizable?

If $A=SDS^{-1}$ with $D=\operatorname{diag}(2,-1)$ , what is $A^3$ ?