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I'm working on linear transformation trying to answer :

Let $E$ and $F$ be two vector spaces on $\mathbb{K}$ and $L:E \rightarrow F$ a function. The graph of $L$ is $\mathbb{G}(L)=\{(x,y) \ \in \ E \times F\mid\ y=L(x)\}$.

Prove $L$ is a linear transformation if and only if $\mathbb{G}(L)$ is a sub vector space of$ E \times F$.

I've tried using the definition of a linear transformation and sub vector space but do not manage to proove the result.

Thank you

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    $\begingroup$ Do you know about the vector subspace test? $\endgroup$ – GFauxPas Apr 19 '15 at 13:47
  • $\begingroup$ Hint: show the closure property of a vector space. $\endgroup$ – sashas Apr 19 '15 at 13:51
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Suppose that $G(L)$ is a linear subspace in $E\times F$.

You need to prove that $$L(\alpha x )= \alpha L(x)\\ L(u+v)=L(u)+L(v).$$ Take any two points $(x_i,y_i)\in E\times F$ belonging to $G(L)$; in other words, $L(x_i)=y_i$, $i=1,2$. By definition of linear subspace you have $(x_1+x_2,y_1+y_2)\in G(L)$. Therefore, by the definition of graph you get $L(x_1+x_2)=y_1+y_2=L(x_1)+L(x_2)$. Similarly, for any constant $\alpha$ you get $(\alpha x_1,\alpha y_1)\in G(L)$ and therefore $L(\alpha x_1)=\alpha y_1 = \alpha L(x_1)$. These two results allow us to conclude that $L$ is linear.

The proof in the other direction is done likewise.

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Hint: $$ (u,L(u))+(v,L(v))= (u+v,L(u)+L(v))=(u+v,L(u+v)) $$ The first identity come from the canonical definition of a product vector space, the second from the definition of $\mathbb{G}(L)$ and from being a vector space.

The same you have for: $$ c(u,L(u))=(cu,cL(u))=(cu,L(cu)) $$

From this you can see that $L$ is linear. The inverse is easy.

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  • $\begingroup$ Actually, this line of reasoning shows that if $L$ is linear, then the graph is a subspace, not the other way around. So "you can see that $L$ linear" is wrong. $\endgroup$ – Thomas Andrews Apr 19 '15 at 14:04
  • $\begingroup$ @Thomas: If The graph is a subspace than the element $(u+v,L(u)+L(v))$ is in the graph and it has the form $(x,L(x))$, so it is $(u+v,L(u+v))$ and this implies that $L(u)+L(v)=L(u+v)$ $\endgroup$ – Emilio Novati Apr 19 '15 at 16:25
  • $\begingroup$ It's true, but that wasn't the argument you made. $\endgroup$ – Thomas Andrews Apr 19 '15 at 16:34
  • $\begingroup$ Fundamentally, you have to state the property of the graph: If $(x,y),(x,z)\in G(L)$ then $y=z$. Then you know that $(cu,cL(u)),(cu,L(cu))\in G(L)$ so $cL(u)=L(cu)$. $\endgroup$ – Thomas Andrews Apr 19 '15 at 16:37
  • $\begingroup$ Thomas. I given a hint. So the detailed argument was left open on purpose. I suppose that it can be suggested by the formula. :) $\endgroup$ – Emilio Novati Apr 19 '15 at 16:54

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