My Math Forum Non-Parallel Eigenvectors with the same Eigenvalue

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 November 17th, 2013, 08:10 AM #1 Newbie   Joined: Sep 2012 Posts: 14 Thanks: 1 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.
November 17th, 2013, 08:12 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: $\left[ \begin{array}{ccc} 3 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right]$ 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: [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.

 November 17th, 2013, 05:20 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Non-Parallel Eigenvectors with the same Eigenvalue The definition of "eigenvalue" and "eigenvector" is that "$\lambda$ is an eigenvalue of A if and only if there exist a non-zero vector v such that $Av= \lambda v$" and "v is an eigenvector corresponding to eigenvalue $\lambda$ of linear transformation, A, if and only if $Av= \lambda v$". If $\lambda= 1$ is an eigenvalue for $\begin{bmatrix}3 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$, then there exist vector $\begin{bmatrix}x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}3=&1=&-1 \\ -1=&1=&1 \\ 1=&1=&1 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}3x+ y- z \\ -x+ y+ z \\ x+ y+ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ z\end{bmatrix}$. 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<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 $\lambda= 2$ is an eigenvalue for $\begin{bmatrix}3 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$, then there exist vector $\begin{bmatrix}x \\ y \\ z \end{bmatrix}$ such that $\begin{bmatrix}3=&1=&-1 \\ -1=&1=&1 \\ 1=&1=&1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix}3x+ y- z \\ -x+ y+ z \\ x+ y+ z\end{bmatrix}= \begin{bmatrix}2x \\ 2y \\ 2z\end{bmatrix}$ 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>= + <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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