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 (1 − λ)(1 − λ) − 2 × 2 = 0.
Multiply out the brackets, and 2 × 2 = 4.
Like terms together, the highest power first: λ² − 2λ − 3 = 0.
It factors, as in the factoring lesson.
So λ = −1 or λ = 3.
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: 1 − (−1) = 2 and 1 − (−1) = 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 = −1 is (−1, 1).
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
3 Eigenvector for λ = 3
We take 3 off the diagonal: 1 − 3 = −2 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 add 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).
So: λ = −1: (−1, 1), and λ = 3: (1, 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, 1): (1, −1).
Now λ = 3.
That is 3 times (1, 1): (3, 3).
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 = 3.
Last: check.