$$ \begin{bmatrix} 1 & 1 & 1\\ 0 & -1 & -2\\ 0 & 0 & 1\\ \end{bmatrix}\\ $$ Since $P$ is upper triangular, the eigenvalues are $1$, $1$ and $-1$.
So the eigenspace of $E(-1)$ is necessarily one-dimensional.
Calculations give that an eigenvector for the eigenvalue $1$ is $(0, 1, -1)$. It does not have two distinct eigenvectors.
Have I missed something?
Help is greatly appreciated.
On
$P$ is diagonalizable $\iff$ $P=V.D.V^{-1}$ where $V$ is the transpose of the EigenVectors matrix, $D$ is the diagonal matrix of eigenvalues and $\det(V)\neq0$ (non-singular).
Eigenvalues associated with $P$ are $(-1,1,1)$ and Eigenvectors $\{(-1, 2, 0), (0, -1, 1), (1, 0, 0)\}$. Note that eigenvalue $\lambda=1$ has two associated vectors
Then transpose Eigenvectors matrix is:
$V^{T} = \begin{pmatrix}-1&0& 1 \\ 2&-1&0 \\ 0&1&0\end{pmatrix}$ whichs is non-singular.
$D = V^{-1}PV = \begin{pmatrix}-1&0&0 \\ 0&1&0 \\ 0&0&1\end{pmatrix}$ as you can see eigenvalues are located in the diagonal
Then we know that $P$ can be expressed as $P = VDV^{-1}$
I don't know what “calculations” you're referring to, but the eigenspace associated with the eigenvalue $1$ is $E(1)=\operatorname{Ker}(P-I_3)$, and $$P-I_3=\begin{pmatrix}0&1&1\\0&-2&-2\\0&0&0\end{pmatrix}$$ the rank of which is clearly $1$ hence, by the rank–nullity theorem, $$\dim E(1)=3-1=2=\text{multiplicity of the eigenvalue $1$}.$$