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)(−λ) − 2 × 2 = 0.
Multiply out the brackets, and 2 × 2 = 4.
Like terms together, the highest power first: λ² − 3λ − 4 = 0.
It factors, as in the factoring lesson.
So λ = −1 or λ = 4.
2 Eigenvector for λ = −1
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take −1 off the diagonal: 3 − (−1) = 4 and 0 − (−1) = 1.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x + y/2 = 0.
We want x alone.
To remove + y/2, we take away y/2 on both sides.
They cancel on the left.
So x = −y/2.
We pick y = 2 to keep whole numbers.
So an eigenvector for λ = −1 is (−1, 2).
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
3 Eigenvector for λ = 4
We take 4 off the diagonal: 3 − 4 = −1 and 0 − 4 = −4.
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − 2y = 0.
We want x alone.
To remove − 2y, we add 2y on both sides.
They cancel on the left.
So x = 2y.
We pick y = 1.
So an eigenvector for λ = 4 is (2, 1).
So: λ = −1: (−1, 2), and λ = 4: (2, 1).
Any number times them, except 0, works too.
4 Check
Let's check: A v must be λ times v.
That is −1 times (−1, 2): (1, −2).
Now λ = 4.
That is 4 times (2, 1): (8, 4).
A only stretches each one, by its λ.
So the answer is right!
Let's look back.
First: find the eigenvalues.
Then: eigenvector for λ = −1.
Then: eigenvector for λ = 4.
Last: check.