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We have a matrix $A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \right)$.

How do you find a diagonal matrix $D$ and an orthogonal matrix $P$ so that:

$D=P^tAP$?

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    Someone else trying to do this may be able to follow it easily from this MIT OCW video: http://www.youtube.com/watch?v=cOUTpqlX-Xs2014-02-17

2 Answers 2

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$A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \right)$.

You know $D = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right)$.

Now, $Ax = \lambda x$

x = 2 (Case 1) $Ax = 2x$ $Ax - 2x = 0$ $(A - 2I)x = 0$ $A -2I= \left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{array} \right)$ $\left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{array} \right)x=0$ Take it to echelon form you get

$\left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{array} \right)x=0$ Thus, if x = (a b c)'

a = b = c Similarly, do for the rest.

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$A$ is symmetric, then you can find an orthonormal basis of ${\bf R}^3$ with eigenvectors. Solve $\det(A-tI)=0$, find the proper values, and for each one the eigenspace. If one of the spaces has dimension 2 you have to apply Gram-Schmidt to get an orthonormal basis, and the other basis will have one vector that you can normalize. Then the union of the two bases is an orthonormal basis (eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal). Put this basis as columns of $P$..

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    Some1, you're not helping. Can you find the eigenvalues? Can you find a basis for each eigenspace? Can you use Gram-Schmidt to get an orthonormal basis for each eigenspace? Can you make a matrix whose columns are the eigenvectors? It's not enough to say you don't understand - you have to show us what you *do* understand, and then we can help you over the rough spots.2012-02-04