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I am having trouble understanding some of the question. I am able to do the computations.

The question asks to find the Eigenvalues and the corresponding Eigenvectors of a matrix $A$, verify that the Eigenvectors are orthogonal, and then compute matrix multiplication.

I have found the Eigenvalues - $\lambda_1 =2$, $\lambda_2=4$, $\lambda_3=6$ - and corresponding Eigenvectors - $\boldsymbol{x_1}$, $\boldsymbol{x_2}$, $\boldsymbol{x_3}$ - of the matrix

$A = \begin{pmatrix} 3 & 0 & -1 \\ 0 & 6 & 0 \\ -1 & 0 & 3 \end{pmatrix}$ and is $P = \left( \frac{\boldsymbol{x_1}}{|\boldsymbol{x_1}|}, \frac{\boldsymbol{x_2}}{|\boldsymbol{x_2}|}, \frac{\boldsymbol{x_3}}{|\boldsymbol{x_3}|} \right)$.

I have verified that $\boldsymbol{x_1}$, $\boldsymbol{x_2}$, $\boldsymbol{x_3}$ are orthogonal.

My question is: how does $P$ represent a matrix and what significance does the fact that the three Eigenvectors are orthogonal have to the question?

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  • $\begingroup$ Have you learnt about a theorem regarding diagonalising real symmetric matrices (search for the "spectral theorem")? And $P$ is a $3\times 3$ matrix with the vectors you have written as its three columns (in that order). $\endgroup$ Mar 6, 2019 at 9:45
  • $\begingroup$ @MinusOne-Twelfth I don't think I have heard of the Spectral theorem, but I will look into it. $\endgroup$
    – Gurjinder
    Mar 7, 2019 at 22:02

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$P$ is the $3 \times 3$ matrix in which each column is a normalized eigenvector of $A$.

The significance of the eigenvectors being orthogonal is that the off-diagonal entries of $P^TP$ will be zero because $x_i . x_j = 0$ if $i \ne j$. And the diagonal entries of $P^TP$ will all be $1$ because the columns of $P$ have been normalized. So $P^TP=I$ i.e. $P$ is an orthogonal matrix.

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The eigenvalue equaiton states $$\boldsymbol{A}\boldsymbol{v}_i=\lambda_i \boldsymbol{v}_i \quad \forall i=1,2,3$$ or in matrix notation $$\boldsymbol{A}[\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3]=[\boldsymbol{v}_1,\boldsymbol{v}_2,\boldsymbol{v}_3]\text{diag}[\lambda_1,\lambda_2,\lambda_3] $$ $$\boldsymbol{AV}=\boldsymbol{V\Lambda},$$ in which $\boldsymbol{V}$ is the matrix containing the normalized eigenvectors as columns (in your question $\boldsymbol{P}$) and $\boldsymbol{\Lambda}$ is a diagonal matrix containing the eigenvalues of $\boldsymbol{A}$. As all your eigenvalues are distinct we know that we can choose the eigenvectors such that he matrix is a orthogonal matrix such that $\boldsymbol{V}^{-1} = \boldsymbol{V}^T$. Solving the previous equation for $\boldsymbol{\Lambda}$ results in

$$\boldsymbol{\Lambda} = \boldsymbol{V}^{-1}\boldsymbol{AV}=\boldsymbol{V}^{T}\boldsymbol{AV}.$$

Hence, we know that $\boldsymbol{V}^T\boldsymbol{AV}$ is nothing than a diagonal matrix $\boldsymbol{\Lambda}$.

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