Eigenvalues and Eigenvectors

Stage 13 of 23 Strand 4 of 5 5 lessons

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

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Eigenvalues and Eigenvectors

The directions a matrix only stretches.

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

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A4−12131117×=

A sends the column (3, 1) to (11, 7), which points in a different direction.

A4−1211133×=

But (1, 1) comes back as (3, 3): the same direction, stretched by a factor of 3.

the matrix acts like one number
v is an eigenvector, its eigenvalue

A vector that A only stretches is an eigenvector: A v = λ v, and λ is its eigenvalue.

A4−1211224×=

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

A4−1212266×=

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

reverses the direction
and then det A = 0
collapses it to the origin

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

Now you

A522242×

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

A223133×

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

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λI, since a number cannot be taken from a matrix

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

with v not the zero column
an inverse would force v = 0
so has no inverse

A − λI sends a nonzero vector to zero, so it has no inverse and its determinant is 0.

A − λI4 − λ−121 − λ

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

the characteristic polynomial

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

or

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

A4−121

Check them: they add to 5, the trace 4 + 1, and multiply to 6, which is det A.

Now you

A − λI4 − λ−121 − λ

What does A − λI look like?

A1423

What is the characteristic polynomial of this matrix?

Finding an Eigenvector

Each eigenvalue leaves one direction behind.

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

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A4−1213I30031−12−2−=

Subtract 3I from A: 3 comes off each entry of the main diagonal.

A − 3I1−12−2xy00×=

An eigenvector for λ = 3 is a nonzero vector that A − 3I sends to zero, so multiply it out.

x − y = 0the second row is twice the first2x − 2y = 0y = x

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

A4−1211133×=

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

A4−1212I20022−12−1−=

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

A4−1211224×=

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

Now you

A0121

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

A − -1I3232

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

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P1112D3002

The eigenvectors are the columns of P, (1, 1) beside (1, 2), and D holds their eigenvalues in that same order, 3 then 2.

A4−121P1112A P3234×=

A P applies A to each column of P, so each column comes out multiplied by its own eigenvalue.

P1112D3002P D3234×=

P D scales those same columns by 3 and by 2, and gives the same matrix.

A P = P Dmultiply on the left by P⁻¹

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

P⁻¹2−1−11A P3234D3002×=

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

A2102

P needs two independent columns. This A has λ = 2 twice but one eigenvector direction, (1, 0), so no P can be built.

Now you

A2102

What stops a matrix from being diagonalized this way?

A4−121P1112D3002

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

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the inner pair is I

Square P D P⁻¹: the inner P⁻¹ P is I, which leaves D² in the middle.

⋯
true for every whole n

The same cancellation happens at every join, so Aⁿ = P Dⁿ P⁻¹ for every n.

D3002D⁵2430032→

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

P1112D⁵2430032P⁻¹2−1−11××

So A⁵ takes two matrix products, not four: P times D⁵, then times P⁻¹.

A⁵454−211422−179

Working it through gives A⁵ = (454 −211; 422 −179) without multiplying A by itself.

Mⁿstartx₀y₀xₙyₙ×=

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

Now you

D3002D⁵2430032→

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

D2004

D = (2 0; 0 4). What is D²?

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