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 (1 − λ)(−1 − λ) − 2 × 4 = 0.
3 Multiply out
Every term of the first bracket times every term of the second.
And 2 × 4 = 8.
Like terms together, the highest power first: λ² − 9 = 0.
Now we solve it, as in the quadratic equations lesson.
4 Get λ² alone
There is no λ term, so we get λ² by itself.
To remove the − 9, we add 9 on both sides.
0 + 9 = 9.
Now we have λ² = 9.
5 Square root of both sides
λ squared is 9.
So λ is the square root of 9, or minus it.
Minus times minus is plus, so the negative one works too.
3 times 3 is 9.
So λ = 3 or λ = −3.
6 Check
Let's check.
The eigenvalues always add up to the diagonal of A.
−3 + 3 = 0, and the diagonal adds up to 0.
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: get λ² alone.
Then: square root of both sides.
Last: check.