Eigenvalues of [1, 2; 2, 1] Step by Step

Step by step

The question

Question: find the eigenvalues of A.

A = [1, 2; 2, 1]

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 = [1 − λ, 2; 2, 1 − 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) = |1 − λ, 2; 2, 1 − λ|

So (1 − λ)(1 − λ) − 2 × 2 = 0.

(1 − λ)(1 − λ) − 2 × 2 = 0

3 Multiply out

Every term of the first bracket times every term of the second.

And 2 × 2 = 4.

1 − λ − λ + λ² − 4 = 0

Like terms together, the highest power first: λ² − 2λ − 3 = 0.

λ² − 2λ − 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 −2.

? × ? = −3

? + ? = −2

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.

× is − → one +, one −; bigger one −

We list the pairs that make 3, smallest first.

We take the smaller from the bigger.

We need 2.

1 and 3 give 2.

Yes!

With the signs: 1 and −3.

1 × 3: 3 − 1 = 2 ✓

So it is (λ + 1)(λ − 3) = 0.

(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.

A × B = 0 → A = 0 or B = 0

We make each bracket 0.

Two small equations.

λ + 1 = 0 or λ − 3 = 0

In λ + 1 = 0, we take away 1 on both sides.

In λ − 3 = 0, we add 3 on both sides.

λ + 1 − 1 = 0 − 1 or λ − 3 + 3 = 0 + 3

So λ = −1 or λ = 3.

λ = −1 or λ = 3

6 Check

Let's check.

The eigenvalues always add up to the diagonal of A.

−1 + 3 = 2, and the diagonal adds up to 2.

λs added: −1 + 3 = 2, diagonal: 1 + 1 = 2

And they multiply to det A: −1 × 3 = −3, and det A is −3.

Both match, so the answer is right!

λs multiplied: −1 × 3 = −3, det A = 1 × 1 − 2 × 2 = −3

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.

Answer