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We know that the dot product with an unit vector gives back the projection in the direction of that unit vector. This means that the mapping $$x \mapsto e(e \cdot x)$$ Is the linear projection to $e$. If we write it out component-wise, we get that $$x_i =\sum_j e_i (e_j x_j)=\sum_j e_i e_j x_j$$ We can notice that the $(i,j)$-th component of the projection matrix is $e_i e_j$.
In your case: $e=\frac{1}{\sqrt{2}} [1,1]^t$, which means that $e_i e_j=\frac{1}{2}$.
The definition of $f$ tells us that $U$ is its image and $W$ its kernel. Moreover, the restriction of $f$ to $U$ is the identity map. If we assume that both $U$ and $W$ are nontrivial, then with a suitable choice of basis the matrix of $f$ is $\operatorname{diag}(1,0)$, i.e., $$A = B\begin{bmatrix}1&0\\0&0\end{bmatrix}B^{-1}$$ for some invertible matrix $B$.
The condition $A(1,1)^T=(1,1)^T$ tells us that $(1,1)^T\in U$. Taking that as the basis for $U$, $$B = \begin{bmatrix}1&*\\1&*\end{bmatrix}.$$ Now, since $V=U\oplus W$, choose any vector that’s not a multiple of $(1,1)^T$ to generate $\ker f$. That vector goes in place of the *s above. For an orthogonal projection, pick a vector orthogonal to $(1,1)^T$.
A similar argument can be made in terms of eigenvalues. The definition of $f$ tells us that the possible eigenvalues of $A$ are $1$ and $0$, and that $(1,1)^T$ is an eigenvector with eigenvalue $1$. $A$ is therefore either the identity matrix (trivial kernel) or similar to $\operatorname{diag}(1,0)$. All that’s left is to choose a vector in the kernel of $f$.
If you allow $U$ or $W$ to be trivial, that opens up the possibilities $A=0$ and $A=I$. We can reject the first since we know of at least one vector that’s not mapped to $0$, but with $W=\{0\}$ we could have the identity matrix.