Let $C(X)$ be the column space of $X \in M_{n \times p}$. Starting off the proof of $C(X) \cup C(X)^{\perp} = \mathbb{R}^n$

Question

Let $C(X)$ be the column space of $X \in M_{n \times p}$. Prove or disprove the following statement: Every vector in $\mathbb{R}^n$ is in either $C(X)$ or $C(X)^{\perp}$ or both.

I interpret this as $C(X) \cup C(X)^{\perp} \overset{?}{=} \mathbb{R}^n$.

I am using the definition $C(X) = $ span of columns of $X$.

Clearly $C(X) \cup C(X)^{\perp} \subset \mathbb{R}^n$ is obvious.

Now let $y \in \mathbb{R}^n$.

This is the part where I'm stuck. Any suggestions on starting this off?

Answer 1

Think of a simple counter-example: $$X = \pmatrix{1&0\\0&0} \in \mathbb R^{2\times 2}; \qquad y = \pmatrix{1\\1}$$ We have $C(X) = \{(x,0) \mid x\in\mathbb R\}$ and $C(X)^\perp = \{(0,x) \mid x\in\mathbb R\}$

Answer 2

If $U_1$ and $U_2$ are subspaces of the vector space $V$, then $U_1\cup U_2$ is a subspace if and only if either $U_1\subseteq U_2$ or $U_2\subseteq U_1$

Proof. Suppose $U_2\not\subseteq U_1$ and let $y\in U_2$, $y\notin U_1$. Let $x\in U_1$; if $U_1\cup U_2$ is a subspace, then $x+y\in U_1\cup U_2$. However $x+y\in U_1$ implies $y\in U_1$, which is false. Thus $x+y\in U_2$ and so $x\in U_2$. The reverse implication is obvious.

So, in order that $C(X)\cup C(X)^\perp=\mathbb{R}^n$ we need either $C(X)\subseteq C(X)^\perp$ or $C(X)^\perp\subseteq C(X)$, that is, that $X$ either has rank $0$ or $n$.