In this exercise, we aim to show that all projections onto finite-dimensional subspaces are symmetric linear transformations, and that the only eigenvalues they can possess are 0 and 1. Let $V$ be an inner product space and $U$ be a finite-dimensional subspace of $V$. We define the projection $T: V \to V$ by $T(v) = \text{proj}_U v$ for all $v \in V$.
$\textbf{Part (a): Symmetric Linear Transformation}$ Initially, I tried to demonstrate this equality by writing $T(v) = U U^T v$ and $T(w) = U U^T w$ as a matrix representation of the projection. Then, calculated using the definition $\langle T(v), w \rangle = \langle v, T(w) \rangle$ to find that both sides are equal. However, I believe there might be another way to approach this by using the definition directly
We will now consider a more abstract approach. Start with the geometric definition of the projection and then use the properties of inner products to arrive at the same result.
Let's begin by expressing $T(v) = v - v'$, where $v'$ is orthogonal to the subspace $U$, and $v - v'$ is a vector in $U$.
Thus, we have:
$$ T(v) = v - v' $$ $$ T(w) = w - w' $$
Then we get:
$$ \langle T(v), w \rangle = \langle v - v', w \rangle = \langle v, w \rangle - \langle v', w \rangle $$
And:
$$ \langle v, T(w) \rangle = \langle v, w - w' \rangle = \langle v, w \rangle - \langle v, w' \rangle $$
It would hold to show that $\langle v, w' \rangle$ and $\langle v', w \rangle$ are equal to zero. However, we can only make this assumption for vectors that are already in the subspace $V$, i.e., $U$. Then $\langle v, w' \rangle = 0$ indeed, because $v \in U$ and $w' \in U^\perp$. Here is where I get confused; this seems like just a special case where it works. It is not obvious that if $u, w \in V$, the inner product would be zero. To show that $T$ is a symmetric linear transformation, we have to prove that $\forall v,w \in V$ ,$ \langle v, T(w) \rangle = \langle T(v), w \rangle$
Your need to show $<v',w> = <v,w'>$.
To prove it, we show $<v,w'> = <v',w'>$, and $<v',w> = <v',w'>$.
$<v,w'> = <v',w'>$ because $<v-v',w'>=0$, $w'$ is orthogonal to $U$ and $v-v'\in U$.
The proof $<v',w> = <v',w'>$ is similar.
Remark: since you tagged your question under linear algebra, I assume $V$ is finite-dimensional. If this is not the case, this proof doesn't work, and you have to use the fact $T$ is a projection onto a finite-dimensional space.