Diagonalising: method and condition

To diagonalise a matrix is to find a viewpoint — a basis — in which it becomes diagonal, and hence trivial to handle.

The idea

If we have a basis made entirely of eigenvectors, then in that basis the matrix merely stretches each axis independently:

        arbitrary basis                basis of eigenvectors

        [ 3   1 ]                          [ 4   0 ]
        [ 2   2 ]                          [ 0   1 ]

    the axes get mixed              each axis is simply stretched

The formula

A = P D P⁻¹

where:

D = DIAGONAL matrix of the eigenvalues
P = change-of-basis matrix: its COLUMNS are the eigenvectors,
    in the SAME ORDER as the eigenvalues in D

On the previous example (λ = 4 with (1 ; 1), λ = 1 with (1 ; -2)):

P = [ 1   1 ]        D = [ 4   0 ]
    [ 1  -2 ]            [ 0   1 ]

Order is the only real source of error: the first column of P must be an eigenvector of the first eigenvalue of D.

The diagonalisability condition

A is diagonalisable  <=>  there is a BASIS of eigenvectors
                     <=>  the dim E(λ) add up to n

In practice two criteria almost always suffice:

1. n DISTINCT eigenvalues     ->  diagonalisable, guaranteed
2. real SYMMETRIC matrix      ->  diagonalisable, guaranteed
                                  (even in an orthonormal basis)

The first criterion settles the example above: two distinct eigenvalues in dimension 2, and we are done.

When it fails

The problematic case is a repeated eigenvalue without enough eigenvectors:

A = [ 1  1 ]      det(A - λI) = (1-λ)²   ->  λ = 1, multiplicity 2
    [ 0  1 ]

E(1) : (A - I)X = 0  ->  [ 0  1 ] [x]   [0]     ->  y = 0
                         [ 0  0 ] [y] = [0]

E(1) = Vect( (1 ; 0) )      dimension 1  <  2 = multiplicity

There is only one eigendirection instead of two: no basis can be formed. This matrix — a shear — is not diagonalisable. Jordan reduction is then used to put it in an almost-diagonal form.

The full method, in four steps

1. compute the characteristic polynomial  det(A - λI)
2. find its roots  ->  the eigenvalues
3. for each λ, solve (A - λI)X = 0  ->  the eigenvectors
4. if n independent eigenvectors are obtained:
      P = their coordinates as columns,  D = the λ in the same order

A possible final check: verify A P = P D, quicker than computing P⁻¹.

Summary

• Diagonalising means writing A = P D P⁻¹ in a basis of eigenvectors.
• D holds the eigenvalues, P the eigenvectors as columns, in the same order.
• Diagonalisable ⟺ there is a basis of eigenvectors.
• n distinct eigenvalues → diagonalisable; real symmetric matrix → diagonalisable.
• It can fail if a repeated eigenvalue lacks eigenvectors (e.g. the shear [[1, 1], [0, 1]]).
• Quick check: A P = P D.