Worked examples
The question
Question: find an eigenvector of A for λ = 3.
A = [2, 1; 1, 2]
An eigenvector is a vector that A only stretches: A times v is λ times v.
The zero vector does not count.
eigenvector: A v = λ v, v not 0
1 Eigenvector for λ = 3
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take 3 off the diagonal: 2 − 3 = −1 and 2 − 3 = −1.
A − 3I = [−1, 1; 1, −1]
→ [1, −1; −1, 1]
→ [1, −1; 0, 0]
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − y = 0.
x − y = 0
We want x alone.
To remove − y, we add y on both sides.
x − y + y = y
They cancel on the left.
So x = y.
x = y
We pick y = 1.
So an eigenvector for λ = 3 is (1, 1).
v = [1; 1]
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
Careful: the zero vector does not count
So: λ = 3: (1, 1).
Any number times them, except 0, works too.
= λ = 3: [1; 1]
2 Check
Let's check: A v must be λ times v.
= [2, 1; 1, 2] × [1; 1]
= [2 × 1 + 1 × 1; 1 × 1 + 2 × 1]
= [3; 3]
That is 3 times (1, 1): (3, 3).
= 3 × [1; 1]
A only stretches it, by its λ.
So the answer is right!
Let's look back.
First: eigenvector for λ = 3.
Last: check.
▶ Watch it solved
The question
Question: find the eigenvectors of A.
A = [2, 1; 1, 2]
An eigenvector is a vector that A only stretches: A times v is λ times v.
The zero vector does not count.
eigenvector: A v = λ v, v not 0
1 Find the eigenvalues
As in the eigenvalue lesson: take λ off the diagonal of A.
A − λI = [2 − λ, 1; 1, 2 − λ]
Its determinant, down minus up, must be 0.
So (2 − λ)(2 − λ) − 1 × 1 = 0.
(2 − λ)(2 − λ) − 1 × 1 = 0
Multiply out the brackets, and 1 × 1 = 1.
4 − 2λ − 2λ + λ2 − 1 = 0
Like terms together, the highest power first: λ² − 4λ + 3 = 0.
λ2 − 4λ + 3 = 0
It factors, as in the factoring lesson.
(λ − 1)(λ − 3) = 0
So λ = 1 or λ = 3.
λ = 1 or λ = 3
2 Eigenvector for λ = 1
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take 1 off the diagonal: 2 − 1 = 1 and 2 − 1 = 1.
A − 1I = [1, 1; 1, 1]
→ [1, 1; 0, 0]
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x + y = 0.
x + y = 0
We want x alone.
To remove + y, we take away y on both sides.
x + y − y = −y
They cancel on the left.
So x = −y.
x = −y
We pick y = 1.
So an eigenvector for λ = 1 is (−1, 1).
v = [−1; 1]
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
Careful: the zero vector does not count
3 Eigenvector for λ = 3
We take 3 off the diagonal: 2 − 3 = −1 and 2 − 3 = −1.
A − 3I = [−1, 1; 1, −1]
→ [1, −1; −1, 1]
→ [1, −1; 0, 0]
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − y = 0.
x − y = 0
We want x alone.
To remove − y, we add y on both sides.
x − y + y = y
They cancel on the left.
So x = y.
x = y
We pick y = 1.
So an eigenvector for λ = 3 is (1, 1).
v = [1; 1]
So: λ = 1: (−1, 1), and λ = 3: (1, 1).
Any number times them, except 0, works too.
= λ = 1: [−1; 1], λ = 3: [1; 1]
4 Check
Let's check: A v must be λ times v.
= [2, 1; 1, 2] × [−1; 1]
= [2 × (−1) + 1 × 1; 1 × (−1) + 2 × 1]
= [−1; 1]
That is 1 times (−1, 1).
= 1 × [−1; 1]
Now λ = 3.
= [2, 1; 1, 2] × [1; 1]
= [2 × 1 + 1 × 1; 1 × 1 + 2 × 1]
= [3; 3]
That is 3 times (1, 1): (3, 3).
= 3 × [1; 1]
A only stretches each one, by its λ.
So the answer is right!
Let's look back.
First: find the eigenvalues.
Then: eigenvector for λ = 1.
Then: eigenvector for λ = 3.
Last: check.
Answer = λ = 1: [−1; 1], λ = 3: [1; 1]
▶ Watch it solved
The question
Question: find the eigenvectors of A.
A = [3, 2; 2, 0]
An eigenvector is a vector that A only stretches: A times v is λ times v.
The zero vector does not count.
eigenvector: A v = λ v, v not 0
1 Find the eigenvalues
As in the eigenvalue lesson: take λ off the diagonal of A.
A − λI = [3 − λ, 2; 2, −λ]
Its determinant, down minus up, must be 0.
So (3 − λ)(−λ) − 2 × 2 = 0.
(3 − λ)(−λ) − 2 × 2 = 0
Multiply out the brackets, and 2 × 2 = 4.
−3λ + λ2 − 4 = 0
Like terms together, the highest power first: λ² − 3λ − 4 = 0.
λ2 − 3λ − 4 = 0
It factors, as in the factoring lesson.
(λ + 1)(λ − 4) = 0
So λ = −1 or λ = 4.
λ = −1 or λ = 4
2 Eigenvector for λ = −1
An eigenvector for this λ is a vector that A − λI sends to 0: its null space.
We take −1 off the diagonal: 3 − (−1) = 4 and 0 − (−1) = 1.
A − (−1)I = [4, 2; 2, 1]
→ [1, 1 over 2; 2, 1]
→ [1, 1 over 2; 0, 0]
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x + y/2 = 0.
x + y2 = 0
We want x alone.
To remove + y/2, we take away y/2 on both sides.
x + y2 − y2 = −y2
They cancel on the left.
So x = −y/2.
x = −y2
We pick y = 2 to keep whole numbers.
So an eigenvector for λ = −1 is (−1, 2).
v = [−1; 2]
Careful!
Every number for the free letter works, except 0.
The zero vector is never an eigenvector.
Careful: the zero vector does not count
3 Eigenvector for λ = 4
We take 4 off the diagonal: 3 − 4 = −1 and 0 − 4 = −4.
A − 4I = [−1, 2; 2, −4]
→ [1, −2; 2, −4]
→ [1, −2; 0, 0]
A row of 0s: the letter y is free.
That is why there are eigenvectors.
Row 1 as an equation: x − 2y = 0.
x − 2y = 0
We want x alone.
To remove − 2y, we add 2y on both sides.
x − 2y + 2y = 2y
They cancel on the left.
So x = 2y.
x = 2y
We pick y = 1.
So an eigenvector for λ = 4 is (2, 1).
v = [2; 1]
So: λ = −1: (−1, 2), and λ = 4: (2, 1).
Any number times them, except 0, works too.
= λ = −1: [−1; 2], λ = 4: [2; 1]
4 Check
Let's check: A v must be λ times v.
= [3, 2; 2, 0] × [−1; 2]
= [3 × (−1) + 2 × 2; 2 × (−1) + 0 × 2]
= [1; −2]
That is −1 times (−1, 2): (1, −2).
= −1 × [−1; 2]
Now λ = 4.
= [3, 2; 2, 0] × [2; 1]
= [3 × 2 + 2 × 1; 2 × 2 + 0 × 1]
= [8; 4]
That is 4 times (2, 1): (8, 4).
= 4 × [2; 1]
A only stretches each one, by its λ.
So the answer is right!
Let's look back.
First: find the eigenvalues.
Then: eigenvector for λ = −1.
Then: eigenvector for λ = 4.
Last: check.
Answer = λ = −1: [−1; 2], λ = 4: [2; 1]
▶ Watch it solved
