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Eigenvectors - Linear Algebra

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QuantumJG:
Determine the Eigenvalue of A if:

A =
2 2 12 5 21 2 2and 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 12 4 2= 0
1 2 11 2 10 0 0= 0
0 0 0so 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 12 -2 2= 0
1 2 -51 0 -10 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.

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