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November 17th, 2013, 09:10 AM   #1
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Non-Parallel Eigenvectors with the same Eigenvalue

Hello,
I have the following matrix:

\left[
\begin{array}{ccc}
3 & 1 & -1 \\
-1 & 1 & 1 \\
1 & 1 & 1 \\
\end{array}
\right]

Using the characteristic equation, I have found that the eigenvalues are 1 and 2.

I am then asked to find two non-parallel eigenvectors with eigenvalue = 2.

I have found that one one eigenvector is: \left[
\begin{array}{c}
1 \\
0 \\
1 \\
\end{array}
\right]

How would I go about finding the other non-parallel eigenvector with eigenvalue=2?

Any help would be greatly appreciated.
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November 17th, 2013, 09:12 AM   #2
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Re: Non-Parallel Eigenvectors with the same Eigenvalue

Quote:
Originally Posted by TomCadwallader
Hello,
I apologise, I didn't include the latex tags. Here it is formatted correctly:

I have been given the following equation:



Using the characteristic equation, I have found that the eigenvalues are 1 and 2.

I am then asked to find two non-parallel eigenvectors with eigenvalue = 2.

I have found that one one eigenvector is: [latex\left[
\begin{array}{c}
1 \\
0 \\
1 \\
\end{array}
\right]][/latex]

How would I go about finding the other non-parallel eigenvector with eigenvalue=2?

Any help would be greatly appreciated.
TomCadwallader is offline  
November 17th, 2013, 06:20 PM   #3
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Re: Non-Parallel Eigenvectors with the same Eigenvalue

The definition of "eigenvalue" and "eigenvector" is that " is an eigenvalue of A if and only if there exist a non-zero vector v such that " and "v is an eigenvector corresponding to eigenvalue of linear transformation, A, if and only if ".

If is an eigenvalue for , then there exist vector such that
.

That is equivalent to the three equations 3x+ y- z= x, -x+ y+ z= y, and x+ y+ z= z. Those then reduce to 2x+ y- z= 0, -x+ z= 0, and x+ y= 0. From -x+ z= 0 we get z= x. From x+ y= 0 we get y= -x. Putting those into 2x+ y- z= 0 we have 2x- x- x= 0 which is true for all x. Because y= -x and z= x, any eigenvector corresponding to eigenvalue 1 must of the form <x, -x, x>= x<1, -1, 1>. The set of all eigenvectors corresponding to eigenvalue 1 make up the one dimensional vector space spanned by <1, -1, 1>.

If is an eigenvalue for , then there exist vector such that


That is equivalent to the three equations 3x+ y- z= 2x, -x+ y+ z= 2y, and x+ y+ z= 2z. Those then reduce to x+ y- z= 0, -x- y+ z= 0, and x+ y- z= 0. Those three equations are all equivalent! The first and third are identical and the second is just the first multiplied by -1. Any x, y, z that satisfy them must have z= x+ y so such an eigenvector is of the form < x, y, x+ y>= <x, 0, x>+ <0, y, y>= x<1, 0, 1>+ y<0, 1, 1>.

Now do you see the two non-parallel eigenvectors corresponding to eigenvalue 2?

(Of course, this can only happen because eigenvalue 2 has "algebraic multiplicity" 2 (2 is a double root of the characteristic polynomial). The fact that the eigenspace has dimension 2 means it has "geometric multiplicity" 2. The geometric multiplicity of an eigenvalue must be less than or equal to its algebraic multiplicity.)
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