Can eigenvalues be interpreted as coordinates in a vector space of projection operators? Consider the vector space $\mathbb R^n$. Let $T : \mathbb R^n \to \mathbb R^n$ be a linear operator, and let $\mathbf T$ be its matrix representation with respect to the standard basis in $\mathbb R^n$. Suppose that the eigendecomposition of $\mathbf T$ is
\begin{align}
\mathbf T &= \mathbf Q \mathbf \Lambda \mathbf Q^{-1} \\
&= \lambda_1 \mathbf v_1 \mathbf u_1^T + \lambda_2 \mathbf v_2 \mathbf u_2^T + \cdots + \lambda_n \mathbf v_n \mathbf u_n^T
\end{align}
where $\lambda_i$ is the $i$th eigenvalue of $\mathbf{T}$, $\mathbf{v}_i$ is the $i$th column in $\mathbf Q$, and $\mathbf{u}_i^T$ is the $i$th row in $\mathbf Q^{-1}$. The outer products $\mathbf v_i \mathbf u_i^T$ represent "scaled" projection operators, such that if each of them was replaced with $\mathbf v_i \mathbf u_i^T / \mathbf u_i^T \mathbf v_i$, then they would be true projection operators.
I am not sure if the set of all projection operators on $\mathbb R^n$ forms a vector space, since this set is not a subspace of the vector space of all linear operators on $\mathbb R^n$ (not closed under scalar multiplication). However, assuming that the set of all projection operators is indeed a vector space, wouldn't the outer products $\mathbf v_1 \mathbf u_1^T,\mathbf v_2 \mathbf u_2^T,\dots,\mathbf v_n \mathbf u_n^T$ form a basis for this vector space? Furthermore, since $\mathbf T$ is expressed as a linear combination of basis vectors, then wouldn't the eigenvalues be coordinates in this vector space? Essentially, what I am asking is: do linear operators on $\mathbb R^n$ live in a vector space of projections, and are they "located" by their eigenvalues in this vector space?
 A: First, the inner product $\mathbf{u}^\top_i \mathbf{v}_i$ is automatically $1$, by the vectors' construction. Think about how we compute $Q^{-1}Q$: by taking dot products of rows of $Q^{-1}$ and columns of $Q$. This implies that
$$\mathbf{u}^\top_i \mathbf{v}_j = I_{ij} = \delta_{ij},$$
i.e. $1$ when $i = j$ and $0$ otherwise. So, the outer products $\mathbf{v}_i\mathbf{u}_i^\top$ are already projections.
But, as for your main question, no, projections are not a vector space, or at least, not obviously a vector space. As you pointed out, they are not a subspace of the space of linear operators. Indeed, your argument starts with an arbitrary diagonalisable operator $\mathbf{T}$, and produces a linear combination of projections. Since not every diagonalisable operator is a projection, this shows that the projections are not a subspace.
If you equip different operations, the set of projections might become a vector space, but if you keep the same addition and scalar multiplication operations that you've been using, then they are definitely not a vector space.
What you have shown is that the projections span the linear operators on $\Bbb{R}^n$. Well, to be fair, not quite: the assumption of an eigendecomposition implies $\mathbf{T}$ is diagonalisable, so you've shown that every diagonalisable operator lies in the span of the projections. You then need another argument for operators that aren't diagonalisable.
You could argue, for example, that if you put any non-diagonalisable argument into Jordan Normal Form, you can add a diagonal (and hence diagonalisable) matrix to it in order to make the matrix diagonalisable. All you need to do is add numbers to the diagonal of the Jordan blocks to make all the eigenvalues distinct. This will produce an upper-triangular matrix with distinct eigenvalues, which must be diagonalisable. This makes every non-diagonalisable operator a difference of two diagonalisable operators, both of which are in the span of the projections. Thus, the projections span every operator.
