# Matrix Calculations Linear Algrebra 3x3 PDP^T

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.

New contributor
Alexis Maldonado is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• 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). Jun 23 at 7:31
• 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? Jun 23 at 8:03
• 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! Jun 23 at 8:07

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.