Orthogonally Diagonalize the matricies The Question given was "Orthogonally diagonalize the matrices, giving an orthogonal matrix P and diagonal matrix D."
I was given eigenvalues -3 and 15
and Matrix A=
$$
\begin{bmatrix}
5 & 8 & -4 \\
8 & 5 & -4 \\
-4 & -4 & -1 
\end{bmatrix}
$$
So far I've done the identity M-I(eigenvalue) with eigenvalue -3 and have gotten
$$
\begin{bmatrix}
8 & 8 & -4 \\
8 & 8 & -4 \\
-4 & -4 & 2 
\end{bmatrix}
$$
Reduced too
$$
\begin{bmatrix}
1 & 1 & -.5 \\
0 & 0 & 0 \\
0 & 0 & 0 
\end{bmatrix}
$$
I'm struggling with what to do after this. I looked at Slader but whoever did this question did it way differently and extremely confused about how he produced his numbers.
 A: The given matrix 
$$ \begin{bmatrix}
5 & 8 & -4 \\ 
5 & 5 & -4 \\ -4 & -4 & -1
\end{bmatrix}
,$$
has eigenvalues $-3$ and two other complex eigenvalues , so I suspect you must have miss-wrote one of the entries of the matrix. Since the matrix is $3\times 3$, by theorem it must have $3$ eigenvalues. So one of the eigenvalues $-3$ or $15$ is repeated twice. You must go over and look which one is repeated. If $-3$ is the one that is repeated , then again by theorem, your diagonal matrix will be 
$$ 
\begin{bmatrix}
-3 & 0 & 0\\
0& -3 & 0\\
0 & 0 & 15
\end{bmatrix}
.$$
Similarly, if your repeated eigenvalue is $15$ the diagonal matrix will be 
$$
\begin{bmatrix}
-3 & 0 & 0\\
0& 15 & 0\\
0 & 0 & 15
\end{bmatrix}.$$

For the eigenvalues part , you correctly reduced the matrix. From there 
\begin{gather}
\begin{bmatrix}
1& 1 & -5 \\ 0 & 0 & 0 \\ 0 & 0 & 0
\end{bmatrix} \implies a+ b -5c =0
\end{gather}
The way you want to approach this is to set two of the variables as free variables ,i.e., 
$$ b=t \ \ , c = s \ \ \implies a = -t + 5s ,$$
then you want to express $a$ as a linear combination of $t$ and $s$ ; 
$$ a = t \begin{pmatrix} 
-1 \\ 1 \\ 0
\end{pmatrix} + s
\begin{pmatrix}
5 \\ 0 \\ 1
\end{pmatrix}.$$
Therefore your two eigenvectors corresponding to $\lambda = -3$ are $v_{1} = (-1 \ \ 1 \ \ 0)^{T}$ and $v_{2} = (5 \ \ 0 \ \ 1)^{T}$. Now since you want to have an orthogonal matrix, you must normalize the eigenvectors. To do this, we divide each entry of the eigenvectors by their respective eigenvector norm giving 
$$ v_{1} = \begin{pmatrix} 
\frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0
\end{pmatrix} \quad \text{and} \quad v_{2} = \begin{pmatrix}
\frac{5}{\sqrt{26}} \\ 0 \\ \frac{1}{\sqrt{26}}
\end{pmatrix}.$$
Your final orthogonal matrix $P$ will have its first and second column the normalized eigenvectors $v_{1}$ and $v_{2}$ respectively. I'll let you do the third eigenvector $v_{3}$.
