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This question already has an answer here:


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

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marked as duplicate by vonbrand, Brandon Carter, Jim, Martin Argerami, Dominic Michaelis Apr 24 '13 at 4:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – vadim123 Apr 24 '13 at 0:57
  • $\begingroup$ @vadim123 See my edit. ^_^ $\endgroup$ – Trancot Apr 24 '13 at 1:04
  • $\begingroup$ 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. $\endgroup$ – Trancot Apr 24 '13 at 1:47
  • $\begingroup$ Yes, actually, it is. The proper place to put this is as an answer to your question. $\endgroup$ – Michael Grant Apr 24 '13 at 3:09
  • $\begingroup$ 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. $\endgroup$ – Trancot Apr 24 '13 at 3:15
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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 $R(T) \subseteq W$?

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  • $\begingroup$ Yes. Just look again. Maybe you'll see what I'm saying. $\endgroup$ – Trancot Apr 24 '13 at 2:06
  • $\begingroup$ @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. $\endgroup$ – Ink Apr 24 '13 at 2:15
  • $\begingroup$ Could my professor just be test us to see if we know that the rank-nullity theorem holds? $\endgroup$ – Trancot Apr 24 '13 at 2:22
  • $\begingroup$ @Trancot What is the question your professor asked you? $\endgroup$ – Ink Apr 24 '13 at 2:26
  • $\begingroup$ "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 $R(T)=W'$." $\endgroup$ – Trancot Apr 24 '13 at 2:33

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