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If dimV = m , dimU = n , F : V -> U : linear dim(ImF)=r then there exists bases of V and U s.t matrix representation of F has the form $$ \pmatrix{I_r & 0\\0 & 0} $$ (where Ir is r-square identity matrix)

step(i) :

I found that basis of kerF has m-r elts that is, {W1,W2,...,W(m-r)} and dimension of V is m.

step(ii) :

So I want to set basis of V as {V1,V2,...,V(r),W1,W2,...,W(m-r)}

for this process, I learned some theorem about this, but I can't remember it. I want to know what theorem need from step(i) to step(ii)

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In many linear algebra texts, you should be able to find a statement of the form

Let $V$ be a vector space of dimension $n$. If $S = \{v_1,\dots,v_k\}$ is a linearly independent set in $V$, then $S$ can be extended to a basis of $V$. That is, there exist vectors $v_{k + 1}, \dots, v_n$ such that $\{v_1,\dots,v_n\}$ is a basis of $V$.

I don't think this theorem is usually given a name. However, it is often presented as a corollary to the dimension theorem. I am confident that the dimension theorem is proven in Friedberg Insel and Spence, and I suspect that this corollary is presented shortly thereafter.

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    $\begingroup$ I have heard somewhere Extension theorem. :) $\endgroup$ – srijan May 1 '17 at 15:48
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    $\begingroup$ @srijan I could believe it, but usually "extension theorem" refers to a result about being able to extend the domain of a function from a subset to the entire space. $\endgroup$ – Omnomnomnom May 1 '17 at 15:54
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    $\begingroup$ Ya I am agree with your opnion. More specifically extending the basis of a subspace to a basis of a vector space. $\endgroup$ – srijan May 1 '17 at 15:58
  • $\begingroup$ @Omnomnomnom Thank you so much. And I got one more problem while proving this, I want to prove that each value of V1 s.t F(V1),...F(Vr) is basis of Im(F), and I think maybe I should the theorem that any spanning set with r elements is basis of Im(F), am I right? $\endgroup$ – 최선웅 May 1 '17 at 17:04
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    $\begingroup$ That's not necessary here. Of course, $\{F(v_1),\dots,F(v_r), F(w_1),\dots,F(w_{m-r})\}$ is a spanning set for $Im(F)$. Since $F(v_1) = \cdots = F(v_r) = 0$, it suffices to show that $F(w_r), \dots, F(w_{m-r})$ are linearly independent. To show that these vectors are linearly independent, note that if they were not linearly independent, then we could necessarily find another element of $\ker(F)$ from the span of $w_r,\dots,w_{m-r}$. $\endgroup$ – Omnomnomnom May 1 '17 at 17:29

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