Question

Orthogonally diagonalize the​ matrix, giving an orthogonal matrix P and a diagonal matrix D. To save​ time, the eigenvalues are -14 and -5.

-9 -4 21 -4 -9 2 2 2 -6

Answer 1

`P=\left(\begin{array}{ccc}-(2)/(3)&-(2)/(3)&(1)/(3)\\(1)/(√(5))&0&(2)/(√(5))\\-(4)/(3√(5))&(√(5))/(3)&(2)/(3√(5))\end{array}\right)`

Explanation:

It is a result that a matrix
`A` is orthogonally diagonalizable if and only if
`A` is a symmetric matrix. According with the data you provided the matrix should be

`A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)`

We know that its eigenvalues are
`\lambda_(1)=-14, \lambda_(2)=-5`, where
`\lambda_(2)=-5` has multiplicity two.

So if we calculate the corresponding eigenspaces for each eigenvalue we have

`E_{\lambda_(1)=-14}=\langle(-2,-2,1)\rangle`,
`E_{\lambda_(2)=-5}=\langle(1,0,2),(-1,1,0)\rangle.`.

With this in mind we can form the matrices
`P, D` that diagonalizes the matrix
`A` so.

`P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)`

and

`D=\left(\begin{array}{ccc}-14&0&0\\0&-5&0\\0&0&-5\\\end{array}\right)`

Observe that the rows of
`P` are the eigenvectors corresponding to the eigen values.

Now you only need to normalize each row of
`P` dividing by its norm, as a row vector.

The matrix you have to obtain is the matrix shown below