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