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 (5 − λ)(2 − λ) − 1 × (−2) = 0.
Multiply out the brackets, and 1 × (−2) = −2.
Like terms together, the highest power first: λ² − 7λ + 12 = 0.
It factors, as in the factoring lesson.
So λ = 3 or λ = 4.
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: 5 − 3 = 2 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 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).
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: 5 − 4 = 1 and 2 − 4 = −2.
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 = 3: (1, 1), 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 3 times (1, 1): (3, 3).
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 = 3.
Then: eigenvector for λ = 4.
Last: check.