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Let $f:V \rightarrow W$ be a linear transformation. Given bases $\{v_i\}_{1\leq i \leq n}$ and $\{w_j\}_{1\leq j \leq m}$ of V and W, respectively, $f$ has an associated $m \times n$ matrix $A$.

I am having trouble showing the the dimension of the row space equals the rank of $f$. I can't use the fact that the rank of the transpose of a matrix equals the rank of the matrix.

I know that the rank of a matrix is the rank of the column space of $A$ and I know how to show that if I have two bases $B$ and $C$, and a linear transformation $f$ and $A=\mathrm{Mat}_{B,C}(f)$, we have $\mathrm{rank}(f)=\mathrm{rank}(A)$. But I am having trouble establishing the connection with the row space.

I am having trouble showing as well that the row space and the nullspace span $V$. I think I need to show the the sum of the row space and the nullspace equal $V$ but I am not sure how to show that. Does the intersection between the row space and the nullspace correspond to $\{0\}$ ?

Thank you

The issue with the previous answer was that it used the relation that the rank of the transpose of a matrix equals the rank of the matrix, which I am not allowed to use.

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See here: en.wikipedia.org/wiki/Rank_(linear_algebra) –  1015 Feb 25 '13 at 18:55
    
By definition, we know that the rank of a matrix equals the rank of the column space. I can then show that rank of the linear transformation equals rank of the representative matrix of T in the bases of V and W. And then, I can show that rank of the column space equals rank of the row space. Is there perhaps a shorter way? –  Carpediem Feb 25 '13 at 19:10
    
Yes, the rank of $f$ is equal to the (column) rank of $A$ for every matrix $A$ of $f$. Then it mounts to showing that the rank of $A$ is equal to the rank of its transpose. You have three different proofs on the wiki page I linked. Not sure there is a shorter way. –  1015 Feb 25 '13 at 19:14
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It would probably be easier to just prove that the nullspace is the orthogonal complement of the rowspace. In either case, the same idea will easily show that the two spaces have null intersection. –  EuYu Feb 25 '13 at 20:37
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@EuYu: This works when our field is the reals or the complex numbers. –  Manos Feb 25 '13 at 20:42

1 Answer 1

Let $A$ be $m \times n$ over an arbitrary field $F$. Then $\dim(R(A))=\dim(R(A^T))={\rm rank}(A)$ and $\dim(R(A^T))+\dim N(A)=n$.

This is not enough to show that $R(A^T)+N(A) = F^n$. Counterexample: Let $F=\mathbb{Z}_2$ and $A=[1 \, \, \, 1]$. Then $N(A) = <[1 \, \, \, 1]^T>=R(A^T)$ and $R(A^T)+N(A)=N(A) \neq \mathbb{Z}_2 ^ 2$.

If $F=\mathbb{R}$ (or $\mathbb{C}$) we proceed as follows: Take $x \in R(A^T) \cap N(A)$. Then $x = A^T y$ and so $0=Ax=A A^T y \Rightarrow 0=y^T A A^T y = ||A^Ty||_2^2 \Rightarrow ||A^T y||_2=0 \Rightarrow 0=A^Ty=x$ and so $R(A^T) \cap N(A)=0$. Hence $$\dim[ R(A^T) + N(A)]= \dim(R(A^T))+\dim N(A)= n \Rightarrow F^n = R(A^T) + N(A)$$.

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I take it that $R(A)$ and $N(A)$ stand for the row and null spaces of $A$, respectively. No, wait, $R(A)$ must be the range of $A$, or of the associated linear transformation $f$, so it's actually the column space of $A$, and $R(A^t)$ is the row space of $A$. But surely this can be done without introducing the $2$-norm. –  Gerry Myerson Feb 26 '13 at 0:24
    
Yes, you are right about the notation. I am interested in seeing a proof without using the 2-norm. –  Manos Feb 26 '13 at 1:10

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