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 (2 − λ)(2 − λ) − 1 × 1 = 0.
3 Multiply out
Every term of the first bracket times every term of the second.
And 1 × 1 = 1.
Like terms together, the highest power first: λ² − 4λ + 3 = 0.
Now we solve it, as in the quadratic equations lesson.
4 Factor
We need two numbers.
They multiply to 3, and they add to −4.
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 3, smallest first.
We add each pair.
We need 4.
1 and 3 give 4.
Yes!
With the signs: −1 and −3.
So it is (λ − 1)(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.
Two small equations.
In λ − 1 = 0, we add 1 on both sides.
In λ − 3 = 0, we add 3 on both sides.
So λ = 1 or λ = 3.
6 Check
Let's check.
The eigenvalues always add up to the diagonal of A.
1 + 3 = 4, and the diagonal adds up to 4.
And they multiply to det A: 1 × 3 = 3, and det A is 3.
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.