Are orthogonal spaces exhaustive, i.e. is every vector in either the column space or its orthogonal complement?

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Quick question about subspaces, just to make sure I have this straight in my head.

Taking an $n\times k$ matrix X with $rank(X)=k$, is every vector in $\mathbb{R}^n$ in either the column space $C(X)$ (which we know has dimension $k$) or its orthogonal complement the left nullspace $N(X^T)$ (which has dimension $n-k$), or are there some vectors in $\mathbb{R}^n$ which are in neither?

Another way of asking this I suppose: if you union the subspaces $C(X)$ and $N(X^T)$, is the result all of $\mathbb{R}^n$?

Thanks!

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Not quite, but almost.

Consider the function $M:\mathbb{R}^2\to\mathbb{R}^2$ given by the matrix $\pmatrix{1 & 0 \\ 0 & 0 }$. This function is just projection onto the $x$ axis, and so is easy to visualize.

What is the column space (range) of this function? The $x$-axis, obviously. What about the left null space? That is the set of vectors that get mapped to $\mathbb{0}$ by $M^{T}$. Which is, also, fairly obviously, the $y$-axis.

So, is the union of the $x$ and $y$ axes the whole of $\mathbb{R}^{2}$? No, of course not. But, every vector in $\mathbb{R}^{2}$ is expressible as the sum of something on the $x$ axis and something on the $y$-axis. So, the whole space is, not the union, but the direct sum of the column space and the left nullspace.