# 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?

$$X\beta = X_{1,1}\beta_1 + X_{1,2}\beta_2 + \dots + X_{1,p}\beta_{p+q},$$ 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 $$X_2$$.
$$\mathbf{y} = \mathbf{X} \mathbf{\beta} + \epsilon= \begin{bmatrix} \mathbf{X}_1, & \mathbf{X}_2 \end{bmatrix} \begin{bmatrix} \mathbf{\beta}_1 \\ \mathbf{\beta}_2 \end{bmatrix} +\epsilon = \mathbf{X}_1 \mathbf{\beta}_1 + \mathbf{X}_2 \mathbf{\beta}_2 +\epsilon$$
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 $$\beta \in \mathbb{R}^{M}, \beta_1 \in \mathbb{R}^{M_1}, \beta_2 \in \mathbb{R}^{M_2}$$).