Suppose $A$ is a $3$ by $5$ matrix with rank $3$. Suppose $A$ is a $3$ by $5$ matrix with rank $3$. Describe the column space of $A$. 
My thoughts: if the matrix has rank $3$, then it has $3$ independent column vectors. That means the column space is $3$ dimensional while we have that each of the columns are in $\mathbb{R}^3$. 
Intuition tells me I can I say that $C(A)=\mathbb{R^3}$ because I think $3$ vectors is enough? Would that be correct, and is there a proof if so?
 A: The rank of a matrix $A$ is defined to be the dimension of the column space of $A$, which also is the number of pivot columns in the matrix. These pivot columns form a basis for the column space, so we know they are linearly independent. In this case because your matrix has rank $3$, there are $3$ pivot columns. Since we know that a set of $n$ linearly independent vectors in $\mathbb{ℝ}^n$ spans $\mathbb{ℝ}^n$, and in this case we have 3 linearly independent columns in $\mathbb{ℝ}^3$, we can conclude that C(A)= $\mathbb{ℝ}^3$. 
A: Fundamental Theorem of Linear Algebra
A matrix $\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$ induces for fundamental subspaces:
$$
\begin{align}
%
  \mathbf{C}^{n} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
  \mathbf{C}^{m} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
    \color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}
%
\end{align}
$$

Here the matrix 
$
\mathbf{A} \in \mathbb{C}^{3\times 5}_{3}
$


Row space
The $3$ rows are a span of $\mathbb{C}^{3}$:
$$
  \mathbf{C}^{m=3} = 
    \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} = \text{span} \left\{\,
 \mathbf{A}_{1,:}, \mathbf{A}_{2,:}, \mathbf{A}_{3,:}
\right\}
$$
The null space $\color{red} {\mathcal{N} \left( \mathbf{A}^{*} \right)}$ is trivial.

Column space
There are $3$ fundamental columns in $\mathbf{A}$. Let's say they are columns $i$, $j$, and $k$.
$$
\begin{align}
%
  \mathbf{C}^{n=5} &= 
    \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}
\oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} \\
%
&= \text{span} \left\{\,
 \mathbf{A}_{i}, \mathbf{A}_{j}, \mathbf{A}_{k}
\right\}
\oplus
    \color{red}{\mathcal{N} \left( \mathbf{A} \right)} 
\end{align}
$$
We know little of the null space $\color{red}{\mathcal{N} \left( \mathbf{A} \right)}$ except that it has dimension $5-3=2$. To find a span, we could use $\mathbf{E_{A}R}$ reduction, or Gram-Schmidt orthogonalization.
