Eigenvalues of [4, 0, 0; 1, 2, 0; 3, 5, -1] Step by Step

Step by step

The question

Question: find the eigenvalues of A.

A = [4, 0, 0; 1, 2, 0; 3, 5, −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 = [4 − λ, 0, 0; 1, 2 − λ, 0; 3, 5, −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

A − λI has 0s on one side of the diagonal.

Its determinant is the diagonal multiplied.

det(A − λI) = (4 − λ)(2 − λ)(−1 − λ)

It must be 0.

(4 − λ)(2 − λ)(−1 − λ) = 0

3 Solve

A product is 0 only when one bracket is 0.

a product is 0: one bracket is 0

Each bracket is 0 when λ is its diagonal number: λ = 4, λ = 2 or λ = −1.

λ = 4 or λ = 2 or λ = −1

4 Check

Let's check.

The eigenvalues always add up to the diagonal of A.

4 + 2 + (−1) = 5, and the diagonal adds up to 5.

λs added: 4 + 2 + (−1) = 5, diagonal: 4 + 2 + (−1) = 5

They match, so the answer is right!

Let's look back.

First: take λ off the diagonal.

Then: determinant = 0.

Then: solve.

Last: check.

Answer