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.
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 $v_{3}$.