# Proof rank linear map is equal to the rank of its transformation matrix

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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• Don´t you mean $im(A)=\{y\in\mathbb{R}^m:Ax=y\text{ has a solution}\}$ etc. and $rk(A)=\operatorname{dim}(im(A))$? – Peter Melech Aug 13 at 12:29
• $im(A)=\{y \in \mathbb{R^n}\vert Ax =y$ 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. – Ludwig Aug 13 at 12:37
• $im(A)\subset \mathbb{R}^m$. In your definition $im(A)=\mathbb{R}^n$ since $A\cdot x=y$ always has a solution if $x$ is given – Peter Melech Aug 13 at 12:52
• 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), $im(A) \subset \mathbb{R^m}$. – Ludwig Aug 13 at 13:04
• You should tell us which definition of the rank of a matrix is used. – Paul Frost Aug 13 at 14:01

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 $$\phi_w(w_i) = e_i$$.

• 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. – Ludwig Aug 13 at 16:33
• I edited my answer. You must not choose an arbitary base. – Paul Frost Aug 13 at 17:08
• 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. – Ludwig Aug 14 at 10:30
• Thank you, you are right. I shall correct it. – Paul Frost Aug 14 at 11:11