# Four Fundamental subspaces

Given an $m \times n$ matrix, where $m$ is the number of rows and $n$ is the number of columns.

Four Fundamental Subspaces

1. The row space is $C(A^t)$, a subspace of $\mathbb{R}^n$.
2. The column space is $C(A)$, a subspace of $\mathbb{R}^m$.
3. The nullspace is $N(A)$, a subspace of $\mathbb{R}^n$.
4. The left nullspace is $N(A^t)$, a subspace of $\mathbb{R}^m$. This is our new space.

If the column space is the space that is spanned by the column vectors, why is it that it is a subspace of $\mathbb{R}^m$ and not $\mathbb{R}^n$, since the dimension of the columns should be $n$ instead?

Or am I missing something fundamental here?

Convince yourself with an example. I'll take $$\begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$$ which is a $(2 \times 3)$-matrix, and the columns $$\begin{pmatrix}1 \\ 4 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}, \begin{pmatrix} 3 \\ 6 \end{pmatrix}$$ belong to $\mathbb{R}^2$ rather than $\mathbb{R}^3$.
No, you're misinterpreting the dimensions of a matrix. Given an $m\times n$ matrix (I'm going to stick with lowercase letters), then as you said correctly yourself,
So $n$ is the number of columns, but not their dimension (or size). Each column has $m$ entries in it because there are $m$ rows in the matrix, i.e. $m$ is the "height" of the matrix and thus the "height" of each column. That's why each column is an element of $\mathbb{R}^m$.