Show that $\pmb y = \pmb X \pmb \beta + \pmb e = \pmb X_1 \pmb \beta_1+ \pmb X_2 \pmb \beta_2 + \pmb e$ Say we take $\pmb X \pmb \beta$ and partition the columns of $\pmb X$ so that we have $\pmb X \pmb \beta_1$ and $\pmb X_2 \pmb \beta_2$. Why does:
$\pmb y = \pmb  X \pmb \beta + \pmb e = \pmb X_1 \pmb \beta_1+ \pmb X_2 \pmb \beta_2 + \pmb e$
I can reason through it geometrically. We're finding the the vector that gets as close to $\pmb y$ as possible in the column space of $\pmb X$. Therefore splitting up the matrix isn't going to change $\hat{ \pmb y}$. Furthermore, the coeficients in $\pmb \beta$ will correspond to the the same columns in the split regression because the solution to the original problem is unique.
So  I feel comfortable with why the result is true but I don't know how to prove it beyond that... Is it possible to do this reasonably?
 A: You may write
$ \mathbf{y} = \mathbf{X} \mathbf{\beta} + \epsilon=
\begin{bmatrix}
\mathbf{X}_1, & \mathbf{X}_2
\end{bmatrix}
\begin{bmatrix}
\mathbf{\beta}_1 \\ \mathbf{\beta}_2
\end{bmatrix}
+\epsilon
=
\mathbf{X}_1 \mathbf{\beta}_1 +
\mathbf{X}_2 \mathbf{\beta}_2
+\epsilon
$
where $\mathbf{X}$ is decomposed in its columns, and $\beta$ in its rows (with appropriate dimensions of course, i.e. if $y \in \mathbb{R}^N$, and $\mathbf{X} \in \mathbb{R}^{N \times M}$, then $\mathbf{X}_1 \in \mathbb{R}^{N\times M_1}$ and $\mathbf{X}_2 \in \mathbb{R}^{N\times M_2}$, where $M_1+M_2=M$, and for $\beta \in \mathbb{R}^{M}, \beta_1 \in \mathbb{R}^{M_1}, \beta_2 \in \mathbb{R}^{M_2}$).
A: By the definition of matrix multiplication,
$$
X\beta = X_{1,1}\beta_1 + X_{1,2}\beta_2 + \dots + X_{1,p}\beta_{p+q},
$$
which is exactly equal to
$$
X_1\beta_1 + X_2\beta_2 = X_{1_{1,1}}\beta_{1_1} + X_{1_{1,2}}\beta_{1_2} + \dots + X_{2_{1,q}}\beta_{2_{q}}
$$
where $X_1$ and $X_2$ are submatrices split by column, $\beta_1$ and $\beta_2$ are similar splits of the vector $\beta$, $p$ is the number of columns of $X_1$, and $q$ is the number of columns of $X_2$.
