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 (−2 − λ)(−2 − λ) − 1 × 1 = 0.
Multiply out the brackets, and 1 × 1 = 1.
Like terms together, the highest power first: λ² + 4λ + 3 = 0.
It factors, as in the factoring lesson.
So λ = −3 or λ = −1.
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: −2 − (−3) = 1 and −2 − (−3) = 1.
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.
3 Eigenvector for λ = −1
We take −1 off the diagonal: −2 − (−1) = −1 and −2 − (−1) = −1.
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 add y on both sides.
They cancel on the left.
So x = y.
We pick y = 1.
So an eigenvector for λ = −1 is (1, 1).
So: λ = −3: (−1, 1), and λ = −1: (1, 1).
Any number times them, except 0, works too.
4 Check
Let's check: A v must be λ times v.
That is −3 times (−1, 1): (3, −3).
Now λ = −1.
That is −1 times (1, 1): (−1, −1).
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 = −3.
Then: eigenvector for λ = −1.
Last: check.