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QUESTION

Prove that there exists a $T:V\rightarrow W$ such that $N(T)=V'\subset V$ and $R(T)=W'\subset W$


ATTEMPTED ANSWER

Let $V$ and $W$ be finite-dimensional vector spaces over $F$. Let $A=\{a_1,\dots,a_l\}$ be a basis for $V'\subset V$, let $B=\{a_1,\dots,a_l,b_1,\dots,b_m\}$ be a basis for $V$, let $C=\{c_1,\dots,c_n\}$ be a basis for $W'\subset W$, and let $D=\{c_1,\dots,c_n,d_1,\dots,d_p\}$ be a basis for $W$. Thus $N(T)=V'$ and $R(T)=W'$ means that

\begin{eqnarray} T(a_1)&=&0+\cdots +0\\ T(a_2)&=&0+\cdots +0\\ \vdots\\ T(a_l)&=&0+\cdots +0\\ T(b_1)&=&k_{11}c_1+k_{21}c_2+\cdots+k_{n1}c_n\\ T(b_2)&=&k_{12}c_1+k_{22}c_2+\cdots+k_{n2}c_n\\ \vdots\\ T(b_n)&=&k_{n1}c_1+k_{n1}c_2+\cdots+k_{nm}c_n, \end{eqnarray}

or simply

\begin{eqnarray} \begin{pmatrix} c_1&c_2&\cdots&c_n \end{pmatrix} \begin{pmatrix} 0&\cdots&0&k_{11}&\cdots&k_{n1}\\ \vdots&&\vdots&\vdots&&\vdots\\ 0&\cdots&0&k_{n1}&\cdots&k_{nm} \end{pmatrix}, \end{eqnarray}

which is an $(l+n)\times (l+m)$ matrix. Thus, such a $T$ exists and has the above form.


JUST CURIOUS

As far as I understand, we're dealing with something that looks like this:

enter image description here

Hanul Jeon
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Trancot
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  • If you would like your answer to be a standalone question, I would recommend copying the relevant parts rather than just giving a link, particularly since that link points to your other question rather than what you want us to look at. – vadim123 Apr 24 '13 at 00:57
  • @vadim123 See my edit. ^_^ – Trancot Apr 24 '13 at 01:04
  • It is not a duplicate! It just simply isn't! ^_^ I'm "asking," as the title alludes, if my answer suffices as a proof and if it doesn't for someone to offer to critique. – Trancot Apr 24 '13 at 01:47
  • Yes, actually, it is. The proper place to put this is as an answer to your question. – Michael Grant Apr 24 '13 at 03:09
  • I argue that it isn't because the question is to be seen through the purview of my tags; that is, it isn't a same-topic duplicate. It is a cross-topic duplicate, which isn't really a duplicate. – Trancot Apr 24 '13 at 03:15

1 Answers1

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I do not see the relevance of your attempted proof to the question you pose.

I'm guessing that $N(T)$ is the kernel of $T$ and $R(T)$ is the image of $T$. Let's look at what these sets are. By definition, we have

$$N(T) = \{v \in V : T(v) = 0\}$$

and

$$R(T) = \{ T(v) : v \in V\}$$

From these definitions, can you convince yourself that $N(T) \subseteq V$ and $R(T) \subseteq W$?

Ink
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  • Yes. Just look again. Maybe you'll see what I'm saying. – Trancot Apr 24 '13 at 02:06
  • @Trancot Like in your other question, it's not clear what you are asking! For *every* linear map $T: V \to W$, it's true that $N(T) \subset V$ and $R(T) \subset W$, and I really don't know how to be anymore clear about this. – Ink Apr 24 '13 at 02:15
  • Could my professor just be test us to see if we know that the rank-nullity theorem holds? – Trancot Apr 24 '13 at 02:22
  • @Trancot What is the question your professor asked you? – Ink Apr 24 '13 at 02:26
  • "Let $V$ and $W$ be finite-dimensional vector spaces over $F$ and $V'\subset V$, $W'\subset W$ two subspaces. Prove that if $\dim(V')+\dim(W')=\dim(V)$, then there is a linear map $T:V\rightarrow W$ such that $N(T) = V'$ and $R(T)=W'$." – Trancot Apr 24 '13 at 02:33
  • I see what your question is asking now. But your proof is insufficient. First of all, note that nowhere did you actually use the condition on the dimensions of the subspaces, which is a necessary condition. Also, you chose your coefficients $k_i$ arbitrarily. What do they mean? If I choose all the $k$s to be $0$ then I just have a zero matrix which shows nothing. Lastly, you haven't shown your matrix (Why would you write it with a row vector by the way? That just feels weird to me) to have the necessary image. The image of the matrix could be a _proper_ subspace of the desired image. – EuYu Apr 24 '13 at 02:49
  • @EuYu What suggestion do you have to re-approach? – Trancot Apr 24 '13 at 03:16
  • Read Berci's answer in your other thread. – EuYu Apr 24 '13 at 03:17