P is given as P = $\left(\begin{array}{rrr} 1 & 1 & 1\\ 1 & 0 & -2\\ 1 & -1 & 1 \end{array}\right).$ It is known that P is invertible.
I is a 3x3 identity matrix
Supposed to find the diagonal matrix D for H, from which we take the diagonal entries to be our eigenvalues.
With the above of which we see what is given, we can conclude with the possibility of using the relation: $ HP = PD $
$ P^{-1} HP = D $
Assuming that this is to be solved using pen and paper (without aid of computers), not spending too much time solving (like during an exam) I believe there could be some shortcut that is simpler/shorter than the above triple matrix multiplication. That multiplication involves pretty tedious arithmetical operations involving squareroots, irrational numbers which is timeconsuming and allows very easily for errors to sneak in. I tried doing it, but it seemed like the wrong way to go because of the complexity.
If $P^{-1}HP = D$, then the columns of $P$ are the eigenvectors of $H$.
So, for every column $p_i$, calculate $Hp_i$ and then use $Hp_i = \lambda_i p_i$ to get the corresponding eigenvalue $\lambda_i$.
So, for the first column you get: $$\begin{align} Hp_1 &= (I- 2uu^T)p_1 = p_1 - 2uu^Tp_1 \\ &= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \frac{2}{3} \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \frac{2}{3} \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \\ -1 \end{pmatrix} \end{align}$$ which gives us that the first eigenvalue is $\lambda_1 = -1$.
This way you don't have to calculate larger matrix products and you don't have to invert $P$.
An even faster way is to see that $H = I - 2uu^T$, for real $u$, is a Householder reflection and therefore will have the eigenvalues $\lambda_1 = -1, \lambda_2 = 1, \lambda_3 = 1$.