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Given $$A =\begin{pmatrix} 0 & 3 & 0 \\ 3 & 0 & 4 \\ 0 & 4 & 0\end{pmatrix}$$ find a matrix $P$ such that $P^T A P$ orthogonally diagonalizes $A$. Verify that $P^TAP$ is diagonal.

I'm not sure how to approach this problem. Any help is appreciated. Thanks!

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You want to find the eigenvalues and eigenvectors of $A$. Since $A$ is a real symmetric matrix, the eigenvectors should be orthogonal if the eigenvalues are distinct (which it is in this case): if they were not distinct you might have to use the Gram-Schmidt procedure to make the eigenvectors orthogonal. Divide each by its length, so they are orthonormal. Then form the matrix $P$ whose columns are those orthonormal eigenvectors.

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  • $\begingroup$ Got it! My solution matched the back of the book. Thanks! $\endgroup$
    – youngdev
    Jul 21 '16 at 17:56
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This should be easy even without calculating any eigenvalues or eigenvectors explicitly. Note that $A=5(uv^T+vu^T)$, where $u=(0,1,0)^T$ and $v=\frac15(3,0,4)^T$ are already two orthonormal vectors. So, if you take $w=\frac15(4,0,-3)^T$, then $Q=(u,v,w)$ is real orthogonal and $$ A=5Q\pmatrix{0&1\\ 1&0\\ &&0}Q^T. $$ Thus the problem boils down to the orthogonal diagonalisation of $\pmatrix{0&1\\ 1&0}$. Yet this $2\times2$ matrix geometrically represents the reflection about the line $x=y$. So, it is immediate that $$ A=5Q\pmatrix{\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\ \frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\\ &&1}\pmatrix{1\\ &-1\\ &&0}\pmatrix{\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\ \frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\\ &&1}Q^T. $$ Absorb the factor $5$ into the diagonal matrix in the middle, you are done.

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