# Show that $P^{T}AP$ reduces to the diagonal matrix $\lambda_{i}\delta_{ij}$

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?

• 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). Mar 6, 2019 at 9:45
• @MinusOne-Twelfth I don't think I have heard of the Spectral theorem, but I will look into it. Mar 7, 2019 at 22:02

$$P$$ 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.
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 $$\boldsymbol{\Lambda}$$.