Eigenvector of [1, 4; 2, -1] for λ = -3

Step by step

The question

Question: find an eigenvector of A for λ = −3.

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 = −3

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

We take −3 off the diagonal: 1 − (−3) = 4 and −1 − (−3) = 2.

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

That is why there are eigenvectors.

Row 1 as an equation: x + y = 0.

We want x alone.

To remove + y, we take away y on both sides.

They cancel on the left.

So x = −y.

We pick y = 1.

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

Careful!

Every number for the free letter works, except 0.

The zero vector is never an eigenvector.

So: λ = −3: (−1, 1).

Any number times them, except 0, works too.

2 Check

Let's check: A v must be λ times v.

That is −3 times (−1, 1): (3, −3).

A only stretches it, by its λ.

So the answer is right!

Let's look back.

First: eigenvector for λ = −3.

Last: check.