Question about projection onto subspaces Suppose I have an $n\times n$ (complex) matrix $Q$ viewed as a linear transformation from $\mathbb{C}^{n}$ to $\mathbb{C}^{n}$. I'm struggling to understand the difference between the following two scenarios. For simplicity, let us consider $n = 4$.
Scenario 1: Let $S$ be a subspace of $\mathbb{C}^{4}$ of dimension $d=2$. Let us think of $S$ as the subspace generated by two eigenvectors $v_{1}$ and $v_{2}$ of $Q$.One can construct a projection matrix $P_{S}$. According to this post, this projection should be given by:
$$P_{S} = A(A^{T}A)^{-1}A^{T}$$
where $A = [v_{1} \hspace{0.1cm} v_{2}]$ the a $4\times 2$ matrix. This is a $4\times 4$ matrix.
Scenario 2: Let us write $\mathbb{C}^{4} = S \oplus S^{\perp}$. Since $Q$ is a linear operator on $\mathbb{C}^{4}$, it might be possible to decompose $Q = Q_{1}\oplus Q_{2}$ so that, if $u = u_{1}+u_{2}$ and $u_{1}\in S$ and $u_{2} \in S^{\perp}$:
$$Qu = Q_{1}u_{1}\oplus Q_{2}u_{2}$$
Each $Q_{i}$ should be a $2\times 2$ matrix.
Here is my problem. I would expect that $Q_{1} = QP_{S}$, since $Q_{1}$ should be the restriction of $Q$ to $S$. However, as I stressed before, both $Q$ and $P_{S}$ are $4\times 4$ matrices. So, where is the gap? If $Q_{1}$ is not $QP_{S}$, how can I obtain $Q_{1}$? And finally, why is $P_{S}$ a $4\times 4$ matrix if it projects into a subspace of dimension $2$? (I would expect it to be a $2\times 4$ matrix instead).

I started working on a trivial example before the much better answer appeared, but perhaps this simple concrete example will help guide someone.
You can express tensors in different bases.
Just like I can express the 2D vector at $v = 3 \mathbf{e}_x + 4 \mathbf{e}_y$ in terms of basis vectors for an $x-y$-plane, I can also express it in terms of $v = 5 \mathbf{e}_v$ where $\mathbf{e}_v = \frac{3}{5} \mathbf{e}_x + \frac{4}{5} \mathbf{e}_y$.
When we only pass around matrices with just the coefficients, e.g. $v = [3,4]$ there is an assumption made of what base they are expressed in.
If we want to project another vector, e.g. $u = 5 \mathbf{e}_x + 0 \mathbf{e}_y$ we can express this projection in several ways, in terms of the 2 coefficients of $\mathbf{e}_x$ and $\mathbf{e}_y$, or in terms of the 1 coefficient of $\mathbf{e}_v$.
Working through the computations for the case I made up here, one will get
$$
u_v = \frac{9}{5} \mathbf{e}_x + \frac{12}{5} \mathbf{e}_y \equiv 3 \mathbf{e}_v
$$
Either way to tackle it, it's the same vector, we only need to think of what base we are expressing the coefficients in.
