Eigenvectors of [3, 1; 0, 3]

Step by step

The question

Question: find the eigenvectors of A.

An eigenvector is a vector that A only stretches: A times v is λ times v.

The zero vector does not count.

1 Find the eigenvalues

As in the eigenvalue lesson: take λ off the diagonal of A.

Its determinant, down minus up, must be 0.

So (3 − λ)(3 − λ) − 0 × 1 = 0.

Multiply out the brackets.

Like terms together, the highest power first: λ² − 6λ + 9 = 0.

It factors, as in the factoring lesson.

So λ = 3.

2 Eigenvector for λ = 3

An eigenvector for this λ is a vector that A − λI sends to 0: its null space.

We take 3 off the diagonal: 3 − 3 = 0 and 3 − 3 = 0.

A row of 0s: the letter x is free.

That is why there are eigenvectors.

Row 1 as an equation: y = 0.

So y = 0.

We pick x = 1.

So an eigenvector for λ = 3 is (1, 0).

Careful!

Every number for the free letter works, except 0.

The zero vector is never an eigenvector.

There is only one λ, and only one free letter: every eigenvector is a number times (1, 0).

So: λ = 3: (1, 0).

Any number times them, except 0, works too.

3 Check

Let's check: A v must be λ times v.

That is 3 times (1, 0): (3, 0).

A only stretches it, by its λ.

So the answer is right!

Let's look back.

First: find the eigenvalues.

Then: eigenvector for λ = 3.

Last: check.