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so I'm supposed to find values for PDP^T that equal to an A given of \begin{bmatrix}7&1&1\\1&7&1\\ 1&1&7\end{bmatrix}

I get eigenvalues of 6,6,9 and my matrices are P=\begin{bmatrix}-1&-1&1\\1&0&1\\ 0&1&1\end{bmatrix} D= \begin{bmatrix}6&0&0\\0&6&0\\ 0&0&9\end{bmatrix} P^T=\begin{bmatrix}-1&1&0\\-1&0&1\\ 1&1&1\end{bmatrix}

My final matrix is \begin{bmatrix}21&3&3\\3&9&15\\ 1&1&7\end{bmatrix}but when I multiply it to 1/3 my final values don't match the original.

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    $\begingroup$ The problem is that the formula is $PDP^{-1}$, not $PDP^\top$. However, since $A$ is symmetric (or even just normal, i.e. $AA^\top = A^\top A$) we can make $P$ an orthogonal matrix, meaning that $P^{-1}$ and $P^\top$ are the same thing. The problem is, we cannot do it by shoving any old eigenvectors as columns in $P$! They need to be orthonormal, so orthogonal and unit length. You need to take your basis of eigenvectors and apply Gram-Schmidt (particularly to the first two, but the third needs to be normalised too). $\endgroup$ Jun 23 at 7:31
  • $\begingroup$ Hi, I'm sorry I thought I already did that. Did I make a mistake in finding the vectors that form my original P? Or was it when I converted it to the inverse? $\endgroup$ Jun 23 at 8:03
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    $\begingroup$ The columns of $P$ are linearly independent eigenvectors, which is a good start. That means $A = PDP^{-1}$. But, if you want $P^{-1}=P^\top$, you need the columns to be orthonormal. Currently they are not. None of the columns are length $1$ (the first two are of length$\sqrt 2$, while the third has length $\sqrt 3$). The first two are also not orthogonal, having a dot product of $1$ instead of $0$. Both are problems here! $\endgroup$ Jun 23 at 8:07

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Since matrix $A$ is symmetric so you can hope $A=PDP^T$ instead of $A=PDP^{-1}$ (in case of non-symmetric matrix), but for this preferred choice you need to satisfy $P^T=P^{-1} $ which is possible iff $P$ is an orthogonal matrix i.e. rows (and columns too) are mutually orthogonal with unit norm. You need to divide each row by its norm which will eventually balance the extra fraction $1/6$ in your calculations.

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