Answer:
[tex] P=\left(\begin{array}{ccc}-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\\-\frac{4}{3\sqrt{5}}&\frac{\sqrt{5}}{3}&\frac{2}{3\sqrt{5}}\end{array}\right)[/tex]
Step-by-step explanation:
It is a result that a matrix [tex]A[/tex] is orthogonally diagonalizable if and only if [tex]A[/tex] is a symmetric matrix. According with the data you provided the matrix should be
[tex] A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)[/tex]
We know that its eigenvalues are [tex]\lambda_{1}=-14, \lambda_{2}=-5[/tex], where [tex]\lambda_{2}=-5[/tex] has multiplicity two.
So if we calculate the corresponding eigenspaces for each eigenvalue we have
[tex]E_{\lambda_{1}=-14}=\langle(-2,-2,1)\rangle[/tex],[tex] E_{\lambda_{2}=-5}=\langle(1,0,2),(-1,1,0)\rangle.[/tex].
With this in mind we can form the matrices [tex]P, D[/tex] that diagonalizes the matrix [tex]A[/tex] so.
[tex] P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)[/tex]
and
[tex] D=\left(\begin{array}{ccc}-14&0&0\\0&-5&0\\0&0&-5\\\end{array}\right)[/tex]
Observe that the rows of [tex]P[/tex] are the eigenvectors corresponding to the eigen values.
Now you only need to normalize each row of [tex]P[/tex] dividing by its norm, as a row vector.
The matrix you have to obtain is the matrix shown below
