0
$\begingroup$

Here's a definition on :

The vector space $L(X,Y)$ of linear maps.

Let $L(X,Y)$ be the set of all linear functions $T:X\rightarrow Y$ .Then $L(X,Y)$ is itself a vector space. The linear operations on $L(X,Y)$ are defined in natural way. If $S,T\in L(X,Y)$ and if $a,b\in \mathbb R$ ,then $R=aS+bT$ is the function : $R$x=$aS$x+$bT$x for x $\in $ $X$. An easy verification shows that $R:$$X \rightarrow Y$ is also linear.Hence $R \in L(X,Y)$

As $R=aS+bT$ What I can't understand is that how did the definition directly found function $R$x=$aS$x+$bT$x i.e. I want to know if we know $R$x=$($a$S$+b$T$$)$x how can we directly write $R$x=$aS$x+$bT$x.

What am I missing ? Please help...

$\endgroup$
  • $\begingroup$ $aS+bT$ is, a priori, just a symbol assembled from two scalars and two linear maps. We want this symbol to describe a linear map $R$, so we need to specify $Rx$ for any $x\in X$. We simply define $R(x)$, for any $x\in X$, by the formula $aS(x)+bT(x)$. $\endgroup$ – Olivier Bégassat Sep 19 '14 at 10:28
  • $\begingroup$ @Olivier Bégassat I can't understand what role does choosing such $R$x has to do with the definition .I very confused Please help... $\endgroup$ – spectraa Sep 19 '14 at 10:32
  • $\begingroup$ Could you try to clearly spell out your confusion? I don't understand what it is : "we want to show that $L(X,Y)$ is a linear map" just doesn't make any sense... $\endgroup$ – Olivier Bégassat Sep 19 '14 at 11:35
  • $\begingroup$ @OlivierBégassat ok my confusion is this: the definition is on showing that $L(X,Y)$is a linear map.so that means that we have to show it's elements satisfies those two axioms.now there is an element $R$ of this to be proven vector space ($L(X,Y)$).My problem /confusion is centered around how did we define the $R(x)$ to be $aS(x)+bT(y)$ can we define $R(x)$ any way as we wish or is there some reason to define it this way.... $\endgroup$ – spectraa Sep 19 '14 at 11:48
1
$\begingroup$

I still don't really understand your confusion, but here goes...

At the most basic level, $L(X,Y)$ is a subset of the set $\mathcal{F}(X,Y)$ of functions from $X$ to $Y$. As a general rule, for any set $S$ and any $\mathbb K$-vector space $V$, the set $\mathcal{F}=\mathcal{F}(S,V)$ of functions $S\to V$ is itself a $\mathbb K$-vector space: it inherits a vector space structure from the vector space structure of $V$. There is a law $\widehat{+}$ for adding two functions $$\widehat{+}:\mathcal{F}\times\mathcal{F}\to\mathcal{F}$$ and a law $\widehat{\,.\,}$ for combining a scalar and a function $$\widehat{\,.\,}:\mathbb{K}\times\mathcal{F}\to\mathcal{F}\,.$$ To be precise, given two functions $f,g:S\to V$ and a scalar $\lambda\in\mathbb K$, you can define two new functions $h,h'$ from $S$ to $V$ by the formulas $$ \forall s\in S,\;h(s)=f(s)+g(s)\text{ and }h'(s)=\lambda\,.f(s) $$ where $+$ is the addition in $V$, and $.$ is scalar multiplication in $V$. We then take $h$ as our definition of $f\,\widehat{+}\,g$, and similarly, we take $h'$ as our definition of $\lambda\,\widehat{\,.\,}\,f$. One checks that these operations satisfy the axioms defining a vector space structure.


In the case where $S$ isn't merely a set but a $\mathbb K$-vector space $X$, we can define a subset $L(X,Y)\subset\mathcal F(X,Y)$ consisting of the linear maps $X\to Y$. One may then ask : is this (non-empty) subset a subspace of the vector space $\mathcal F(X,Y)$? The answer is: yes. Given two linear functions $f,g:X\to Y$ and a scalar $\lambda\in\mathbb K$, the functions $f\,\widehat{+}\,g$ and $\lambda\,\widehat{\,.\,}\,f$ are still linear maps from $X\to Y$.

$\endgroup$
  • $\begingroup$ Thanks I just got this concept now . $\endgroup$ – spectraa Sep 19 '14 at 12:13
  • $\begingroup$ @WantTobeAbstract great! $\endgroup$ – Olivier Bégassat Sep 19 '14 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.