My Math Forum Eigenvector of symmetric matrix

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 October 30th, 2012, 11:14 AM #1 Newbie   Joined: Oct 2012 Posts: 15 Thanks: 0 Eigenvector of symmetric matrix the vector (1 0 1)^t is an eigenvector of the symmetric matrix 6 -1 3 -1 7 x 3 x y how can i find x and y and the corresponding eigenvalue l.. I have no idea how to approach this question.. my understanding is that (1 0 1)^T is actually: 1 0 1 rows become columns and all that.. but how i can find the eigenvalue and original vector using this i dont know.. any help is appreciated!
 November 6th, 2012, 06:09 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Eigenvector of symmetric matrix Do you not know what "eigenvalue" and "eigenvector" mean? A vector, v, is an eigenvector of linear transformation A, with eigenvalue $\lambda$, if and only if $Av= \lambda v$. I don't know what you mean by "the original vector". The is only one vector and it is given. Since we are told that $<1, 0, 1>$ is an eigenvector there must be a number $\lambda$ such that $\begin{bmatrix} 6=&-1=&3 \\ -1=&7=&x \\ 3=&x=&y \end{bmatrix}\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix}\lambda \\ 0 \\ \lambda\end{bmatrix}$. Multiplying out that left side, $\begin{bmatrix}9 \\ x- 1 \\ 3+ y \end{bmatrix}= \begin{bmatrix}\lambda \\ 0 \\ \lambda\end{bmatrix}$ which gives the three equations $9= \lambda$, $x- 1= 0$, and $3+ y= \lambda$. That's pretty easy, isn't it?

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