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 (6 − λ)(1 − λ) − 2 × (−2) = 0.
Multiply out the brackets, and 2 × (−2) = −4.
Like terms together, the highest power first: λ² − 7λ + 10 = 0.
It factors, as in the factoring lesson.
So λ = 2 or λ = 5.
2 Eigenvector for λ = 2
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take 2 off the diagonal: 6 − 2 = 4 and 1 − 2 = −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 add 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 = 2 is (1, 2).
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
3 Eigenvector for λ = 5
We take 5 off the diagonal: 6 − 5 = 1 and 1 − 5 = −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 = 5 is (2, 1).
So: λ = 2: (1, 2), and λ = 5: (2, 1).
Any number times them, except 0, works too.
4 Check
Let's check: A v must be λ times v.
That is 2 times (1, 2): (2, 4).
Now λ = 5.
That is 5 times (2, 1): (10, 5).
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 = 2.
Then: eigenvector for λ = 5.
Last: check.