[ALGEBRA] Diagonalisation

This page asks when a matrix has enough independent eigenvectors to form an eigenbasis. In that basis the matrix representation is diagonal, so repeated multiplication or coupled evolution splits into scalar modes, as in matrix powers, linear ODE systems and Markov systems.

Diagonalisation always takes place over a chosen scalar field: the change-of-basis matrix and the diagonal entries must belong to that field. A real matrix may therefore be diagonalisable over $\mathbb C$ but not over $\mathbb R$ .

Independent eigenvectors

[DEFINITION]

Eigenvalue

Let $T:V\to V$ be a linear map over a field $F$ . A scalar $\lambda\in F$ is an eigenvalue of $T$ if there is a non-zero vector $v\in V$ such that

T(v)=\lambda v.

[1]:Linear map
[2]:Eigenvectors for distinct eigenvalues
[3]:A real matrix with complex eigenvalues
[4]:Invertibility and eigenvalues
[5]:Eigenvector
15 more

[DEFINITION]

Eigenvector

If $T(v)=\lambda v$ for a non-zero vector $v$ , then $v$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$ .

[1]:Eigenvalue
[2]:Eigenvectors for distinct eigenvalues
[3]:Eigenbasis
[4]:A real matrix with complex eigenvalues
[5]:An eigenvector of differentiation
7 more

[DEFINITION]

Basis

A set $B$ in a vector space $V$ is a basis for $V$ if it is linearly independent and spans $V$ .

[1]:Span
[2]:Linear independence
[3]:Eigenbasis
[4]:A system with zero initial derivative
[5]:Values on a basis determine a linear map
[6]:Linear transformations
22 more

[THEOREM]

Matrix representation of a linear map

Let $T:V\to W$ be a linear map, where $\dim(V)=n$ and $\dim(W)=m$ . For ordered bases

B_V=(v_1,\ldots,v_n),\qquad B_W=(w_1,\ldots,w_m),

form the coordinate columns

a_j=[T(v_j)]_{B_W}\qquad (j=1,\ldots,n).

Then

A=\begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}

is the unique $m\times n$ matrix satisfying

[T(v)]_{B_W}=A[v]_{B_V}\qquad\text{for every }v\in V.

We call $A$ the matrix representation of $T$ relative to $B_V$ and $B_W$ and may write it as $[T]_{B_V}^{B_W}$ .

With this notation, the subscript names the domain basis and the superscript names the codomain basis, matching the direction $[v]_{B_V}\mapsto[T(v)]_{B_W}$ in the displayed coordinate identity.

[1]:Linear map
[2]:Basis
[3]:Coordinate vector
[4]:Values on a basis determine a linear map
[5]:Diagonalisation
[6]:Characteristic polynomial
[7]:Eigenvalues and eigenvectors
[8]:First-order linear system
9 more

[THEOREM]

Eigenvectors for distinct eigenvalues

Eigenvectors belonging to distinct eigenvalues are linearly independent. In particular, an $n\times n$ matrix with $n$ distinct eigenvalues has $n$ linearly independent eigenvectors.

[1]:Eigenvalue
[2]:Eigenvector
[3]:Linear independence
[4]:Proof
[5]:A triangular matrix with an eigenbasis

[PROOF]

Let $v_1,\ldots,v_n$ be eigenvectors for distinct eigenvalues $\lambda_1,\ldots,\lambda_n$ and suppose

c_1v_1+\cdots+c_nv_n=0.

To isolate one coefficient, fix $k$ and apply

B_k=\prod_{j\neq k}(A-\lambda_jI)

to both sides. The factors commute because each is a polynomial in $A$ . Every term except the $k$ th is killed by one factor, while

B_kv_k=\prod_{j\neq k}(\lambda_k-\lambda_j)v_k.

Thus the transformed relation is

0=c_k\left(\prod_{j\neq k}(\lambda_k-\lambda_j)\right)v_k.

The scalar product is non-zero and $v_k\ne0$ , so $c_k=0$ . Since $k$ was arbitrary, all coefficients vanish.

[1]:Eigenvectors for distinct eigenvalues

[DEFINITION]

Eigenbasis

An eigenbasis for a vector space is a basis whose vectors are eigenvectors of the linear map under consideration.

In an eigenbasis, each coordinate direction is scaled independently. Every vector therefore splits into eigenmodes, which matrix powers and linear ODE systems evolve separately.

[DEFINITION]

Diagonalisation

An $n\times n$ matrix $A$ over a field $F$ is diagonalisable over $F$ if there are an invertible matrix $P\in M_n(F)$ and a diagonal matrix $D\in M_n(F)$ such that

P^{-1}AP=D, \qquad\text{equivalently}\qquad A=PDP^{-1}.

The columns of $P$ are the eigenvectors in the chosen eigenbasis and the corresponding eigenvalues appear on the diagonal of $D$ .

[THEOREM]

When diagonalisation is possible

An $n\times n$ matrix over $F$ is diagonalisable over $F$ if and only if it has $n$ linearly independent eigenvectors in $F^n$ .

This test decides whether matrix powers can be computed in diagonal form.

[PROOF]

If $v_1,\ldots,v_n$ are independent eigenvectors, let

P=\begin{pmatrix}v_1&\cdots&v_n\end{pmatrix}, \qquad D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n).

The matrix $P$ is invertible and applying $A$ to each column gives $AP=PD$ . Hence $P^{-1}AP=D$ .

Conversely, if $P^{-1}AP=D$ , then $AP=PD$ . The columns of $P$ are therefore eigenvectors corresponding to the diagonal entries of $D$ . Since $P$ is invertible, its columns are linearly independent.

[1]:When diagonalisation is possible

Similarity and change of basis

Two square matrices $A$ and $B$ are similar if an invertible matrix $P$ satisfies

B=P^{-1}AP.

Similar matrices represent the same linear operator in different bases, as in a matrix representation, so their entries can differ. They share eigenvalues, determinant, rank, nullity and invertibility. Diagonalisation $A=PDP^{-1}$ is the case in which the matrix in the new basis is diagonal.

Worked examples

[EXAMPLE]

A real matrix with complex eigenvalues

For

A= \begin{pmatrix} 2&-1\\ 1&2 \end{pmatrix},

the characteristic polynomial is

p_A(\lambda)=(2-\lambda)^2+1 =\lambda^2-4\lambda+5.

Its eigenvalues are $\lambda=2+i$ and $\lambda=2-i$ . For $\lambda=2+i$ ,

A-(2+i)I= \begin{pmatrix} -i&-1\\ 1&-i \end{pmatrix},

and solving $(A-(2+i)I)v=0$ gives

v=\alpha\begin{pmatrix}i\\1\end{pmatrix}.

The conjugate eigenvalue has eigenvector $\begin{pmatrix}-i\\1\end{pmatrix}$ . These two eigenvectors are independent over $\mathbb C$ , so the matrix is diagonalisable over $\mathbb C$ although it has no real eigenvalues. Discarding the complex roots of a real matrix would miss this diagonalisation.

[routine]

A repeated eigenvalue needs more care. Its algebraic multiplicity counts how often it repeats as a root of the characteristic polynomial. Its geometric multiplicity is the dimension of its eigenspace, which is never larger.

A repeated root is not itself an obstruction to diagonalisation over $F$ . The obstruction is an eigenspace that is too small, leaving fewer than $n$ independent eigenvectors in $F^n$ . The three repeated-eigenvalue cases compare the possibilities.

[EXAMPLE]

A repeated eigenvalue without enough modes

Let

A= \begin{pmatrix} 3&-1\\ 1&1 \end{pmatrix}.

Then

p_A(\lambda) =(3-\lambda)(1-\lambda)+1 =(\lambda-2)^2.

The only eigenvalue is $2$ . Solving

(A-2I)v= \begin{pmatrix} 1&-1\\ 1&-1 \end{pmatrix}v=0

gives

E_2=\operatorname{span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}.

The algebraic multiplicity is $2$ , but the geometric multiplicity is $\dim(E_2)=1$ . There is only one independent eigenvector, so no eigenbasis of $\mathbb R^2$ exists and $A$ is not diagonalisable over either $\mathbb R$ or $\mathbb C$ .

[routine]

[1]:Eigenspace
[2]:Diagonalisation

[EXAMPLE]

A triangular matrix with an eigenbasis

For

A= \begin{pmatrix} 1&2&3\\ 0&4&5\\ 0&0&6 \end{pmatrix},

the eigenvalues are the diagonal entries $1$ , $4$ and $6$ . They are distinct, so the corresponding eigenvectors are independent. Solving $(A-\lambda I)v=0$ gives $v_1=(1,0,0)^{\mathsf T}$ for $\lambda=1$ . For $\lambda=4$ , the last two equations give $z=0$ and the first gives $3x=2y$ , so choose $v_2=(2,3,0)^{\mathsf T}$ . For $\lambda=6$ , the second equation gives $2y=5z$ ; taking $z=10$ gives $y=25$ and the first equation then gives $x=16$ . Thus one choice is

v_1= \begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad v_2= \begin{pmatrix}2\\3\\0\end{pmatrix}, \qquad v_3= \begin{pmatrix}16\\25\\10\end{pmatrix}.

With

P= \begin{pmatrix} 1&2&16\\ 0&3&25\\ 0&0&10 \end{pmatrix}, \qquad D= \begin{pmatrix} 1&0&0\\ 0&4&0\\ 0&0&6 \end{pmatrix},

the column checks $Av_j=\lambda_jv_j$ give $AP=PD$ and hence

A=PDP^{-1}.

[routine]

Symmetric matrices

Real symmetric matrices are especially well behaved: their eigenvalues are real and their eigenvectors can be chosen mutually perpendicular. This is the spectral theorem, stated here without proof. The orthonormal form makes the change of basis easy to invert, because the inverse of $Q$ is its transpose.

[PROPOSITION]