Eigenvectors - Linear Algebra

[QuantumJG]

Determine the Eigenvalue of A if:

A =

2 2 1

2 5 2

1 2 2

and then determine the Eigenvectors.

We haven't looked at Eigenvalues or Eigenvectors yet! This is just stuff for before we look a it.

So I found the Eigenvalues: (1,7)

so this is what I did:

lamda = 1,

A - lamda*I = 0

1 2 1

2 4 2

= 0

1 2 1

1 2 1

0 0 0

= 0

0 0 0

so let y = s and z = t, s,t ϵ ℝ

(x,y,z) = s(-2,1,0) + t(-1,0,1)

.: Eigenvectors are {s(-2,1,0), t(-1,0,1)} - this spans a plane in R^3

lamda = 7,

A - lamda*I = 0

-5 2 1

2 -2 2

= 0

1 2 -5

1 0 -1

0 1 -2

= 0

0 0  0

(x,y,z) = w(1,2,1), w ϵ ℝ

therefore the Eigenvector is {w(1,2,1)} - this spans a line in R^3

is this how you find the Eigenvectors?

[Mao]

That is correct.

An eigenvalue is defined as , where is a scalar.
This implies , where can be found by , and v is any vector that satisfies the condition.