Worked examples
The question
Question: find the eigenvalues of A.
A = [2, 0; 0, 5]
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.
A v = λ v (v not 0)
λ (lambda): an eigenvalue
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.
(A − λI) v = 0 → det(A − λI) = 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.
A − λI = [2 − λ, 0; 0, 5 − λ]
Careful!
λ is taken away only from the diagonal entries.
The other entries stay as they are.
Careful: λ comes off the diagonal only
2 Determinant = 0
We want the determinant of A − λI to be 0.
det(A − λI) = |2 − λ, 0; 0, 5 − λ|
So (2 − λ)(5 − λ) − 0 × 0 = 0.
(2 − λ)(5 − λ) − 0 × 0 = 0
3 Multiply out
Every term of the first bracket times every term of the second.
10 − 2λ − 5λ + λ2 = 0
Like terms together, the highest power first: λ² − 7λ + 10 = 0.
λ2 − 7λ + 10 = 0
Now we solve it, as in the quadratic equations lesson.
4 Factor
We need two numbers.
They multiply to 10, and they add to −7.
? × ? = 10
? + ? = −7
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 10, smallest first.
We add each pair.
We need 7.
2 and 5 give 7.
Yes!
With the signs: −2 and −5.
1 × 10: 1 + 10 = 11
2 × 5: 2 + 5 = 7 ✓
So it is (λ − 2)(λ − 5) = 0.
(λ − 2)(λ − 5) = 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.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
λ − 2 = 0 or λ − 5 = 0
In λ − 2 = 0, we add 2 on both sides.
In λ − 5 = 0, we add 5 on both sides.
λ − 2 + 2 = 0 + 2 or λ − 5 + 5 = 0 + 5
So λ = 2 or λ = 5.
λ = 2 or λ = 5
6 Check
Let's check.
The eigenvalues always add up to the diagonal of A.
2 + 5 = 7, and the diagonal adds up to 7.
λs added: 2 + 5 = 7, diagonal: 2 + 5 = 7
And they multiply to det A: 2 × 5 = 10, and det A is 10.
Both match, so the answer is right!
λs multiplied: 2 × 5 = 10, det A = 2 × 5 − 0 × 0 = 10
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.
▶ Watch it solved
The question
Question: find the eigenvalues of A.
A = [5, −2; 1, 2]
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.
A v = λ v (v not 0)
λ (lambda): an eigenvalue
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.
(A − λI) v = 0 → det(A − λI) = 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.
A − λI = [5 − λ, −2; 1, 2 − λ]
Careful!
λ is taken away only from the diagonal entries.
The other entries stay as they are.
Careful: λ comes off the diagonal only
2 Determinant = 0
We want the determinant of A − λI to be 0.
det(A − λI) = |5 − λ, −2; 1, 2 − λ|
So (5 − λ)(2 − λ) − 1 × (−2) = 0.
(5 − λ)(2 − λ) − 1 × (−2) = 0
3 Multiply out
Every term of the first bracket times every term of the second.
And 1 × (−2) = −2.
10 − 5λ − 2λ + λ2 + 2 = 0
Like terms together, the highest power first: λ² − 7λ + 12 = 0.
λ2 − 7λ + 12 = 0
Now we solve it, as in the quadratic equations lesson.
4 Factor
We need two numbers.
They multiply to 12, and they add to −7.
? × ? = 12
? + ? = −7
They multiply to a plus number, so the two numbers have the same sign.
They add to a minus number, so both are minus.
× is + → same signs: both −
We list the pairs that make 12, smallest first.
We add each pair.
We need 7.
3 and 4 give 7.
Yes!
With the signs: −3 and −4.
1 × 12: 1 + 12 = 13
2 × 6: 2 + 6 = 8
3 × 4: 3 + 4 = 7 ✓
So it is (λ − 3)(λ − 4) = 0.
(λ − 3)(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.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
Two small equations.
λ − 3 = 0 or λ − 4 = 0
In λ − 3 = 0, we add 3 on both sides.
In λ − 4 = 0, we add 4 on both sides.
λ − 3 + 3 = 0 + 3 or λ − 4 + 4 = 0 + 4
So λ = 3 or λ = 4.
λ = 3 or λ = 4
6 Check
Let's check.
The eigenvalues always add up to the diagonal of A.
3 + 4 = 7, and the diagonal adds up to 7.
λs added: 3 + 4 = 7, diagonal: 5 + 2 = 7
And they multiply to det A: 3 × 4 = 12, and det A is 12.
Both match, so the answer is right!
λs multiplied: 3 × 4 = 12, det A = 5 × 2 − 1 × (−2) = 12
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.
▶ Watch it solved
The question
Question: find the eigenvalues of A.
A = [3, 1; 0, 3]
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.
A v = λ v (v not 0)
λ (lambda): an eigenvalue
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.
(A − λI) v = 0 → det(A − λI) = 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.
A − λI = [3 − λ, 1; 0, 3 − λ]
Careful!
λ is taken away only from the diagonal entries.
The other entries stay as they are.
Careful: λ comes off the diagonal only
2 Determinant = 0
We want the determinant of A − λI to be 0.
det(A − λI) = |3 − λ, 1; 0, 3 − λ|
So (3 − λ)(3 − λ) − 0 × 1 = 0.
(3 − λ)(3 − λ) − 0 × 1 = 0
3 Multiply out
Every term of the first bracket times every term of the second.
9 − 3λ − 3λ + λ2 = 0
Like terms together, the highest power first: λ² − 6λ + 9 = 0.
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.
? × ? = 9
? + ? = −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.
× is + → same signs: both −
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.
1 × 9: 1 + 9 = 10
3 × 3: 3 + 3 = 6 ✓
So it is (λ − 3)(λ − 3) = 0.
(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.
A × B = 0 → A = 0 or B = 0
We make each bracket 0.
The two brackets are the same.
λ − 3 = 0 or λ − 3 = 0
In λ − 3 = 0, we add 3 on both sides.
λ − 3 + 3 = 0 + 3 or λ − 3 + 3 = 0 + 3
Both brackets give the same answer.
So there is only one answer: λ = 3.
λ = 3 is the only eigenvalue: it comes twice.
λ = 3
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.
λs added: 3 + 3 = 6, diagonal: 3 + 3 = 6
And they multiply to det A: 3 × 3 = 9, and det A is 9.
Both match, so the answer is right!
λs multiplied: 3 × 3 = 9, det A = 3 × 3 − 0 × 1 = 9
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.
▶ Watch it solved
