$$A = \begin{bmatrix} 1 & 1 & -\sqrt{2} \\ 1 & 0 & 0 \\ -\sqrt{2} & 0 & 1 \end{bmatrix}$$ Show that $A^{4} - 2A^{3} + A = O$ where $O$ is the $3\times 3$ zero matrix.
I was able to get the eigenvalues -1, $(3+\sqrt{5})/2$, and $(3-\sqrt{5})/2$. And I thought that $A^{4} - 2A^{3} + A$ would be $D^{4} -2D^{3} + D$ for the diagonal matrix D with those eigenvalues, but I don't get the zero matrix when I try that. Isn't $A = PDP^{T}$? Not really sure where to go from here.
We invoke the characteristic polynomial: $$\det(\mathbf{A}-z\mathbf{I})=\begin{vmatrix}1-z&1&-\sqrt{2}\\ 1&-z&0\\ -\sqrt{2}&0&1-z\end{vmatrix}\\=-z(z-1)^2+z-1+2z\\ =-z^3+2z^2+2z-1.$$ Thus, $$\boxed{\boxed{\mathbf{A}^3-2\mathbf{A}^2-2\mathbf{A}+\mathbf{I}=\mathbf{O}}}.$$ Since $\mathbf{A}$ is invertible, you want to prove that $\mathbf{A}$ satisfies $$\mathbf{A}^3-2\mathbf{A}^2+\mathbf{I}=\mathbf{O},$$ which implies $$\mathbf{A}=\mathbf{O},$$ which is not the case.