Eigenvector of [3, 0; 1, 2] for λ = 2

Step by step

The question

Question: find an eigenvector of A for λ = 2.

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

The zero vector does not count.

1 Eigenvector for λ = 2

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

We take 2 off the diagonal: 3 − 2 = 1 and 2 − 2 = 0.

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

That is why there are eigenvectors.

Row 1 as an equation: x = 0.

So x = 0.

We pick y = 1.

So an eigenvector for λ = 2 is (0, 1).

Careful!

Every number for the free letter works, except 0.

The zero vector is never an eigenvector.

So: λ = 2: (0, 1).

Any number times them, except 0, works too.

2 Check

Let's check: A v must be λ times v.

That is 2 times (0, 1): (0, 2).

A only stretches it, by its λ.

So the answer is right!

Let's look back.

First: eigenvector for λ = 2.

Last: check.