Step by step
The question
Question: find the eigenvalues of A.
Some vectors are only stretched by A: A times v is just a number times v.
That number is called an eigenvalue.
We write it λ, lambda.
Move λv across: A − λI, times v, is 0.
A vector that is not 0 goes to 0 only when the determinant is 0.
So the determinant of A − λI must be 0.
1 Take λ off the diagonal
λI has λ down the diagonal and 0s elsewhere.
So A − λI is A with λ taken away from the diagonal entries only.
Careful!
λ is taken away only from the diagonal entries.
The other entries stay as they are.
2 Determinant = 0
We want the determinant of A − λI to be 0.
So (3 − λ)(3 − λ) − 0 × 1 = 0.
3 Multiply out
Every term of the first bracket times every term of the second.
Like terms together, the highest power first: λ² − 6λ + 9 = 0.
Now we solve it, as in the quadratic equations lesson.
4 Factor
We need two numbers.
They multiply to 9, and they add to −6.
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
We list the pairs that make 9, smallest first.
We add each pair.
We need 6.
3 and 3 give 6.
Yes!
With the signs: −3 and −3.
So it is (λ − 3)(λ − 3) = 0.
5 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
We make each bracket 0.
The two brackets are the same.
In λ − 3 = 0, we add 3 on both sides.
Both brackets give the same answer.
So there is only one answer: λ = 3.
λ = 3 is the only eigenvalue: it comes twice.
6 Check
Let's check.
The eigenvalues always add up to the diagonal of A.
3 + 3 = 6, and the diagonal adds up to 6.
And they multiply to det A: 3 × 3 = 9, and det A is 9.
Both match, so the answer is right!
Let's look back.
First: take λ off the diagonal.
Then: determinant = 0.
Then: multiply out.
Then: factor.
Then: one bracket is 0.
Last: check.