Can we say that the columns of the given matrix always lies in its range space Can we say that the columns of the given matrix always lies in its range space. 
For example, suppose we have a square matrix $A$ of order $n\times n$ then can we claim that its columns say $c_1, c_2 \ldots \in R(A)$.
Please help me to clear my doubt. 
Thank you very much.
 A: Yes. There is always a vector that can be chosen to produce a column from $A$:
Let $A$ be an $n \times n$ matrix.  Let $c_0, c_1, \ldots, c_n$ be its columns.
Let $e_1, e_2, \ldots, e_n$ be the (standard) basis vectors.
Then $Ae_1 = c_1$, $Ae_2 = c_2$, and so on.
A: Let's work over $\mathbb R$, the field of real numbers. Also, let's consider the elements of $\mathbb R^m$ als column vectors. If $A$ is a matrix with real entries and $n$ rows and $m$ columns, then $A$ defines a linear map from $\mathbb R^m$ to $\mathbb R^n$ via $\mathbb R^m \ni v \mapsto Av \in \mathbb R^n$. Here, we also consider the elements of $\mathbb R^n$ as column vectors. The range of $A$ is the subspace $A\mathbb R^m \subseteq \mathbb R^n$.
Now, $\mathbb R^m$ contains the unit vectors $e_j = (0,\ldots,0,1,0,\ldots,0)^T$ with $1$ in the $j$-th row and $0$ elsewhere, for $j \in \{1,\ldots,m\}$. Moreover, $Ae_j$ is just the $j$-th column of $A$. So we have that the $j$-th column of $A$ is contained in the range of $A$ for every $j \in \{1,\ldots,n\}$, as desired.
In fact, the columns of $A$ span the range of $A$ as a subspace of $\mathbb R^n$.
