Eigenvalues and Eigenvectors

Stage 13 of 23 Strand 4 of 5 5 lessons

5 illustrated lessons, each teaching the why before the how.

Revise Eigenvalues and Eigenvectors with flashcards →

Jump to a lesson

Eigenvalues and Eigenvectors #

The directions a matrix only stretches.

An eigenvector is a direction a matrix only stretches, and its eigenvalue is how much

A sends the column (3, 1) to (11, 7), which points somewhere else entirely.

But (1, 1) comes back as (3, 3): the same direction, stretched three times.

A column the matrix only stretches is an eigenvector, and A v = λ v says it.

And (1, 2) is doubled, so it is another eigenvector, with eigenvalue 2.

Doubling an eigenvector gives another, so an eigenvector names a direction.

An eigenvalue may be negative, which reverses the direction, or 0, which collapses it.

Now you

A (2; 1) = 6 (2; 1). What is A (4; 2)?

A (1; 1) = 4 (1; 1). What is A (3; 3)?

The Characteristic Polynomial #

det(A − λI) = 0, and its roots are the eigenvalues.

The eigenvalues of a matrix are the roots of det(A − λI) = 0

Bring everything to one side. The λ must become λI before v can be factored out.

A non-zero column sent to zero means that matrix has no inverse, so its det is 0.

Subtracting λI takes λ off each entry of the main diagonal, and nothing else.

Its determinant is a quadratic in λ, called the characteristic polynomial.

Set it to zero and factor: the two eigenvalues of A are 2 and 3.

Check them: they add to the diagonal total 5, and multiply to det A, which is 6.

Now you

What does A − λI look like?

What are the eigenvalues of this matrix?

Finding an Eigenvector #

Each eigenvalue leaves one direction behind.

Each eigenvalue gives a direction, found by solving (A − λI) v = 0

Take the eigenvalue 3 off the diagonal of A and this is what is left.

The eigenvector is whatever this matrix sends to zero, so multiply it out.

Both rows reduce to y = x, and they must: two different conditions would force v = 0.

Every point on y = x works, so take the simplest one: v = (1, 1).

Repeat at λ = 2: both rows of A − 2I read 2x − y = 0, so y = 2x.

That gives v = (1, 2), and multiplying by A doubles it, so its eigenvalue is 2.

Now you

One eigenvalue of A is -1. Which column is its eigenvector?

Why do both rows of A − λI give the same condition?

Diagonalizing a 2 × 2 Matrix #

Eigenvectors into P, eigenvalues down D.

Putting the eigenvectors in the columns of P makes P⁻¹AP the diagonal matrix of eigenvalues

Stand the two eigenvectors up as columns: (1, 1) beside (1, 2).

D carries their eigenvalues on its diagonal, kept in that same order: 3 then 2.

A P works A on each column of P, so each one comes out times its own λ.

P D scales those same columns by 3 and by 2, and lands on the very same matrix.

So A P = P D. Multiply on the right by P⁻¹ and A is written as P D P⁻¹.

Working it out returns D exactly, so those two eigenvectors diagonalize A.

Now you

What stops a matrix from being diagonalized this way?

Which equation defines the diagonalization of A?

Matrix Powers by Diagonalization #

Aⁿ = P Dⁿ P⁻¹, and a power becomes cheap.

Aⁿ = P Dⁿ P⁻¹ turns a matrix power into powers of the numbers on a diagonal

Square P D P⁻¹ and the inner P⁻¹ P collapses, leaving D squared in the middle.

The same collapse happens n times over, so Aⁿ = P Dⁿ P⁻¹ for every n.

A diagonal matrix is easy to raise to a power: each entry is raised on its own, 3⁵ and 2⁵.

So A⁵ is three products, not five: P, then the raised diagonal, then P⁻¹.

Working it through gives A⁵ = (454 −211; 422 −179) with no repeated squaring.

Population models use this: Mⁿ acting on a starting pair gives the pair n years later.

Now you

Why is A⁵ less work through P D⁵ P⁻¹ than by multiplying A five times?

D = (2 0; 0 4). What is D²?

Continue your journey in the app — save your progress