Given the inner product space $\mathbb R^3$ and $U=sp\{(1,0,-1),(0,1,1)\}$, give an example of a normal matrix $B_{3x3}$ such that $B$ isn't diagonal and one of its eigenspaces is $U$.
I'm not quite sure how to start. I already found that $U^\perp = sp\{(-1,1,-1)\}$.
I also know that $\forall v\in \mathbb R^3 \ : \ v=P_U(v) + P_{U^\perp}(v)$ where $P_U(v)$ is the projection of the vector $v$ on the subspace $U$ and $P_{U^\perp}(v)$ is the projection of the vector $v$ on the subspace $U^\perp$.
I tried to construct $B$ using brute force (i.e. guessing and guessing...) but it didn't help me so much.
Let $u_1:=(1,0,-1)^T$, $u_2:=(0,1,1)^T$ and what you found, $u_3:=(-1,1,-1)^T$.
So we are looking for a matrix $M$ such that $$Mu_1=\lambda u_1\\ Mu_2=\lambda u_2\\ Mu_3=\mu u_3$$ where $\lambda\ne\mu$ are (almost) arbitrary. You can calculate with these variables $\lambda,\mu$ and see when the resulting matrix will be diagonal, or you could take concrete values first (for example $\lambda=0$ and $\mu=3$) to make calculations easier.
Now, if you can express the standard basis $e_1,e_2,e_3$ by linear cobinations of $u_1,u_2,u_3$, then you can also express $Me_1,Me_2,Me_3$, which are just the columns of $M$.