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...

  • $\begingroup$ $aS+bT$ 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

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 $X\to Y$.

  • $\begingroup$ 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

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