Rank of linear transformations.

Question

How could I provide a simple proof for the following statments:

(1) $A$ is an $n*m$ matrix, $T_A:F^n \mapsto F^m$ is a linear transformation such that $T_A(v) = A*v\;$ show that $rank(T_A) = rank(A)$

(2) $V,W$ are vector spaces,$\;B,C$ are bases for $V,W$ respectivly.

$T:V \mapsto W$ is a linear transformation

$A = \left [ T \right ]_{C}^{B}\;$

show that $rank(A) = rank(T)$

It's kind of intuitive why this is true, yet I find it hard to put down a simple proof for it.

Answer 1

For (2), you can prove the following lemma before (it's really very simple) :

Lemma : Let $\textsf V$ and $\textsf W$ finite-dimensional vector spaces and $\Phi : \textsf V \to \textsf W$ an isomorphism between them. If $\textsf U$ is a subspace of $\textsf V$, then $\Phi (\textsf U)$ is a subspace for $\textsf W$ and moreover $\dim (\textsf U) = \dim \Phi (\textsf U)$.

Now, the proof for (2) would begin as follows :

$\textbf{(2).}$ Let $\textsf T : \textsf V \to \textsf W$ be a linear transformation from an $n$-dimensional vector space $\textsf V$ to an $m$-dimensional vector space $\textsf W$. Let $\beta$ and $\gamma$ be ordered basis for $\textsf V$ and $\textsf W$, respectively. Then, $\operatorname{rank}(\textsf T) = \operatorname{rank}(A)$, where $A=[\textsf T]_{\beta}^{\gamma}$.

Proof : Suppose $\beta = \{ u_1,u_2,\dots,u_n \}$ and $\gamma = \{ v_1,v_2,\dots,v_m \}$. Let $\phi_\beta : \textsf V \to \textsf F^n$ the unique linear transformation such that $\phi_\beta (u_j) = e_j$ for $j=1,2,\dots,n$ and $\phi_\gamma : \textsf W \to \textsf F^m$ defined by $\phi_\gamma (v_i) = f_i$ for $i=1,2,\dots,m$.

The vectors $e_1,e_2,\dots,e_n$ represents the standard basis for $\textsf F^n$ and $f_1,f_2,\dots,f_m$ for $\textsf F^m$. And it's easy to see also that $\phi_\beta$ and $\phi_\gamma$ are the both isomorphisms, a very important relation between them is that $$\textsf T_A \circ \phi_\beta = \phi_\gamma \circ \textsf T$$ Now, since $\phi_\beta$ is surjective we have that $$\operatorname{im}(\textsf T_A) = \textsf T_A (\textsf F^n ) = \textsf T_A (\phi_\beta (\textsf V)) = \phi_\gamma (\textsf T (\textsf V)) = \phi_\gamma (\operatorname{im}(\textsf T))$$ Since $\operatorname{im}(\textsf T)$ is a subspace of $\textsf W$ and $\phi_\gamma$ an isomorphism, by the lemma, it follows that $$\begin{align} \operatorname{rank} (\textsf T) &= \dim (\operatorname{im} (\textsf T)) \\ &= \dim \phi_\gamma (\operatorname{im}(\textsf T)) \\ &= \dim (\operatorname{im}( \textsf T_A) )\\ &= \operatorname{rank} (\textsf T_A) \end{align}$$ and by (1) we have the desired result.

Answer 2

For 1):

$rank(T) := dim(Im(T))$

Let $B:=\{v_1,..,v_n\}$ be the canonical for $F^n$. Then, $\forall v \in F^n, T(v) \in <T(v_1), ...T(v_n) >$, therefore $Im(T)=<T(v_1),...T(v_n)>$ (the other inclusion is trivial), which of course implies $dim(Im(T))=dim(<T(v_1),...T(v_n)>)$

But $T(v_1), ..., T(v_n)$ are the columns of $A$, so $dim(<T(v_1),...T(v_n)>)=rank(A)$