Equality of rank between linear transformation and its matrix representation $V$ has basis $\{v_1, v_2,..., v_n\}$ and $W$ has basis $\{w_1, w_2,..,w_m\}$. Show that the rank of the linear transform $L:V\to W$ is equal to the rank of the matrix representing $L$ w.r.t. its the bases.
 A: I'll suppose the rank of a linear transformation $L$ is defined as the dimension of its image $L(V)=\{\, L(v)\mid\,v\in V\}$ and that the rank of an $m\times n$ matrix$~A$ is defined as the dimension of the subspace of$~K^m$ (with $K$ the base field, you may take $K=\Bbb R$) spanned by its columns.
First note that if $\{ v_1,\ldots, v_n\}$ is a basis of$~V$, then the vectors $L(v_1),\ldots, L(v_n)\in W$ span $L(V)$, since $L$ respects linear combinations. Also, the matrix $A$ above defines a linear transformation $L_A:K^n\to K^m$ by $L_A(v)=A\cdot v$. Since column$~j$ of $A$ is just $A\cdot e_j$ where $\{e_1,\ldots,e_n\}$ is the standard basis of $K^n$, we have that the subspace spanned by the columns of$~A$ is the subspace of $K^m$ spanned by $L_A(e_j)$ for $j=1,\ldots,n$, which by the above is the image of$~L_A$. So according to the definitions, the rank of $A$ is the rank of$~L_A$.
Finally the bases of the question define isomorphisms $\phi_v:V\to K^n$ (finding coordinates of any vector in $V$ in the basis $\{ v_1,\ldots, v_n\}$) and $\psi_w:K^m\to W$ (using the $m$-tuple of scalars as coordinates in the basis $\{ w_1,\ldots, w_m\}$ to get a vector of$~W$), and the matrix $A$ of$~L$ with respect to these bases is by definition such that $L=\psi_w\circ L_A\circ\phi_v$ (given $v\in V$ find its coordinates by$~\phi_v$, multiply those by$~A$, and interpret the result as a vector of$~W$ via$~\psi_w$). The isomorphism $\psi_w$ then maps the image of $L_A$ to that of $L$, so the two linear transformations have the same rank, which is also the rank of $A$.
A: Suppose $T: ℝ^n → ℝ^m$  defined by $A x = b$ for $A ∈ M_{×}, x ∈ ℝ^n$ and $b ∈ ℝ^m $
We have $A x = x_1 A_{col_1} + x_2 A_{col_2} + ... + x_n A_{col_n}$
Now, suppose {$A_{col_1}, A_{col_2},..., A_{col_r}$} is the maximal linearly independent subset of $A$ columns and $r$ is an arbitrary natural number.
We have: $A x = x_1 A_{col_1} + x_2 A_{col_2} + ... + x_n A_{col_n} = a_1 A_{col_1} + a_2 A_{col_2} + ... + a_r A_{col_r}$
for $a_1, a_2,..., a_r ∈ ℝ$
Therefore, the dimension of the range of $T$ = the rank of $A$.
