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Author Topic: Eigenvectors - Linear Algebra  (Read 1098 times)  Share 

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QuantumJG

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Eigenvectors - Linear Algebra
« on: September 20, 2009, 10:32:15 pm »
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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

(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

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

is this how you find the Eigenvectors?


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Re: Eigenvectors - Linear Algebra
« Reply #1 on: September 21, 2009, 12:09:32 am »
0
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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