I've started reading a book on linear models, and it turns out there's a lot of linear algebra (which I'm a bit weak in). There were two exercises that I think might be simple, but I don't know if I'm doing right.
(1) You are given that $X$ is an $(n \times p)$ matrix and $b ∈ R^p$. Prove that the linear model $y =Xb ∈ R^n$ satisfies the conditions of a vector space.
(2) For a linear model $y = Xb$, is the origin 0 in the column space $C(X)$? Why or why not?
OK, so for (1) I think what I need to do is show that for any two vectors that are in $y$, then the addition of those two vectors are also in $y$ (addition rule). The other condition I think, is that for any scalar $q$ in $R^n$, then $qy$ must also be in $y$. If I have this right, can someone show me how I go about proving this? My current answer seemed a little simple, which is why I doubt it.
For (2), it actually confuses me. I thought the column space referred to all possible linear combinations of the columns of $X$ where the scalars which multiply each column is nonzero. So by definition, I thought you can't have the zero vector be in the column space of $X$? Seems a little weird to me, unless I'm missing something obvious.
I'd really appreciate any help.
1) Let $C(X)$ be the span of the columns of $X$. Take $y_1, y_2$ that satisfies $X\beta_1$ and $X\beta_2$ respectively, so $y_1+y_2=X\beta_1+X\beta_2=X(\beta_1+\beta_2)=X\beta^*$. I.e., there are set of coefficients $\{\beta^*_i=\beta_{1i}+\beta_{2i} \}_{i=1}^p$ such that $\sum_{i=1}^p\beta^*_i\mathrm{x}_i=y_1+y_2$, hence $y_1+y_2 \in {C(X)}$.
scalar multiplication is even easier: $\alpha \in \mathbb{R}$, so $\alpha y= \alpha X\beta=X\alpha\beta$, so the set of the coefficients is given by $\{\alpha\beta_i\}_{i=1}^p$, thus $\alpha y \in C(X)$. Hence, $C(X)$ is a vector subspace.
2) By definition, every vector subspace contains the $0$ vector. In particular, you can take $\alpha=0$ and since $\alpha y\in C(X),\,\, \forall \alpha \in \mathbb{R}$ hence $0 \in C(X)$.