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I want to prove, that the rank of a linear map f: V $\rightarrow$ W is equal to the rank of the transformation matrix A of this linear map.

Let $v$ a Basis of $V$ with length $n$ and $w$ a Basis of $W$ with length m and let $A$ be the $(m$ x $n)$ transformation matrix of the linear map f in relation to $v$ and $w$ with $rk(A)=r$.

Now, let $P$ be an invertible $(n$ x $n)$ matrix, Q an invertible $(m$ x $m)$ matrix. It follows, that $v\cdot P$ is a new Basis of $V$ and $w \cdot Q$ is a new Basis of $W$. The new transformation matrix $B$ of the linear map $f: V \rightarrow W$ in relation to the new bases $v\cdot P$ and $w \cdot Q$ is now given by $B=Q^{-1} \cdot A \cdot P$.

Since $im(A \cdot P)=\{x \in \mathbb{R^n} \vert (A \cdot P)x = y $ has a solution$\}=\{x \in \mathbb{R^n}\vert A \cdot x = y $ has a solution$\}=im(A)$

$\Rightarrow im(Q^{-1} \cdot A \cdot P)=im(Q^{-1}\cdot A)=\{ x \in \mathbb{R^m}\vert (Q^{-1}\cdot A)x=y$ has a solution$\}=\{y \in \mathbb{R^m} \vert Q^{-1}\cdot x = y $ has a solution$, x \in im(A)\}=im(A)$, since $im(Q^{-1})=\mathbb{R^m}$

$\Rightarrow dim (im(f))=rk(f):=r=rk(A)=dim(im(A))=dim(im(B))=rk(B)$ which is therefore not determined by the choice of the bases of $V$ and $W$. $\Box$

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  • $\begingroup$ Don´t you mean $im(A)=\{y\in\mathbb{R}^m:Ax=y\text{ has a solution}\} $ etc. and $rk(A)=\operatorname{dim}(im(A))$? $\endgroup$ – Peter Melech Aug 13 at 12:29
  • $\begingroup$ $im(A)=\{y \in \mathbb{R^n}\vert Ax =y$ has a solution$\}$ is how we definied the image in University. I just checked that. You are right with $dim(im(A))$, $dim(im(B))$. I changed that. $\endgroup$ – Ludwig Aug 13 at 12:37
  • $\begingroup$ $im(A)\subset \mathbb{R}^m$. In your definition $im(A)=\mathbb{R}^n$ since $A\cdot x=y$ always has a solution if $x$ is given $\endgroup$ – Peter Melech Aug 13 at 12:52
  • $\begingroup$ Only $Q^{-1}$ and $P$ are invertible by definition and are therfore of full rank. So $im(Q^{-1})=\mathbb{R^m} \land im(P)=\mathbb{R^n}$. Since $A$ is only of rank $r$ (which I added now in my proof), $im(A) \subset \mathbb{R^m}$. $\endgroup$ – Ludwig Aug 13 at 13:04
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    $\begingroup$ You should tell us which definition of the rank of a matrix is used. $\endgroup$ – Paul Frost Aug 13 at 14:01
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Your proof is not correct. First observe that $im(A)$ can be written simply as $$im(A) = \{ A \cdot x \mid x \in \mathbb R^n\} \subset \mathbb R^m .$$ You correctly show that $im(A \cdot P) = im(A)$. However, you have $$im(Q^{-1} \cdot A \cdot P) = \{ Q^{-1} \cdot ((A \cdot P)\cdot x) = (Q^{-1} \cdot A \cdot P)\cdot x \mid x \in \mathbb R^n \} = \{ Q^{-1} \cdot z \mid z \in im(A \cdot P) = im(A) \} ,$$ but in general the latter differs from $im(A)$. Anyway, it is irrelvant, you only have to compare dimensions.

For an $m\times n$-matrix $A$ let $l_A : \mathbb R^n \to \mathbb R^m$ be the linear map given by $l_A(x) = A \cdot x$. You know that $rk(A) = \dim(im(A)) = \dim (im(l_A))$.

Given a linear map $f : V \to W$ and bases $v =\{v_1,\dots,v_n\}$ of $V$ and $w =\{w_1,\dots,w_m\}$ of $W$, you can form the transformation matrix $A$ of $f$ with respect to $v, w$. Let $\phi_v : V \to \mathbb R^n$ be the linear isomorphism determined by $\phi_v(v_j) = e_j$, where the $e_j$ are the standard basis vectors of $\mathbb R^n$, simalarly $\phi_w : W \to \mathbb R^m$. Then by definition of $A$ we get $\phi_w \circ f \circ (\phi_v)^{-1} = l_A$. This immediately implies $\dim(im(f)) = \dim(im(l_A)) = rk(A)$ because the dimension of linear subspaces is preserved under linear isomorphisms.

Edited on request:

You know that the matrix $A$ is constructed as follows. Since $w$ is a basis of $W$, for each $v_j \in v$ there exists a unique represention $$f(v_j) = \sum_{i=1}^m a_{ij}w_i $$ with $a_{ij} \in \mathbb R$. Then we have $A = (a_{ij})$. What is the purpose of this matrix? Using the above isomorphisms $\phi_v, \phi_w$, we get $$(*) \quad A \cdot \phi_v(x) = l_A(\phi_v(x)) = \phi_w(f(x)) ,$$ i.e. we can reduce $f$ to matrix multiplication.

To verify $(*)$ it suffices to consider $x = v_j$. We get $$A \cdot \phi_v(v_j) = A \cdot e_j = (a_{1j},\dots,a_{mj})^{T} = \sum_{i=1}^m a_{ij}e_i$$ and $$\phi_w(f(v_j)) = \phi_w(\sum_{i=1}^m a_{ij}w_i) = \sum_{i=1}^m a_{ij}\phi_w(w_i) = \sum_{i=1}^m a_{ij}e_i .$$ Here it is essential that $\phi_v(v_j) = e_j$ and $\phi_w(w_i) = e_i$.

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  • $\begingroup$ Thanks a lot for your answer! Is it important to choose the standard basis vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$ or would it change anything in the proof if I would choose some random basis vectors? Just to understand your proof in detail. PS: Sadly I wasn't able to upvote your answer because it says I'm missing reputation. That is why I just ticked it as answered. $\endgroup$ – Ludwig Aug 13 at 16:33
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    $\begingroup$ I edited my answer. You must not choose an arbitary base. $\endgroup$ – Paul Frost Aug 13 at 17:08
  • $\begingroup$ Certainly it must be $\phi(v_j)$ in the 4th line counted from the bottem of your answer. Again, I'm not able to edit because of missing reputation. $\endgroup$ – Ludwig Aug 14 at 10:30
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    $\begingroup$ Thank you, you are right. I shall correct it. $\endgroup$ – Paul Frost Aug 14 at 11:11

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