<aside> 💡 A function $f:V \rightarrow W$ between two vector spaces $V$ and $W$ is said to be a linear transformation if for any two vectors $v_1$ and $v_2$ in the vector space $V$ and for any $c\in\R$(scalar), the following conditions hold: →$f(v_1+v_2)=f(v_1)+f(v_2)$ →$f(cv_1)=cf(v_1)$
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Linear Mappings are Linear Transformations.
<aside> 💡 A function $f:V\rightarrow W$ is one-one(or injective) if $f(v_1)=f(v_2) \implies v_1=v_2$.
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<aside> 💡 A function $f:V\rightarrow W$ is onto(or surjective) if for every $w\in W$ there exists $v\in V$ such that $f(v)=w$.
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<aside> 💡 For a Linear Transformation, being one-one is equivalent to $f(v)=0 \implies v=0$.
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Let us assume $f$ is one-one.($f$ is a linear transformation) Then $f(v_1)=f(v_2)\implies v_1=v_2$.→(1) Since $f$ is a linear transformation, $f(0)=0$ →(2) ($\because f(0+0)=f(0)+f(0) \implies f(0)=f(0)+f(0) \implies f(0)-f(0)=f(0)+f(0)-f(0)(subtract ~~f(0)~~from
bothsides) \implies f(0)=0$ Then if $f(v)=0$, From (2) , $f(v)=f(0)$ → (3) From (1) and (3), $v=0$ Conversely, Let us assume $f(v)=0$,→(1) Then let $f(v_1)=f(v_2)\implies f(v_1-v_2)=0$ →(2)($\because f(v_1)-f(v_2)=0 \implies f(v_1)+f(-v_2)=0 ,$ then by Linear Transformation,$f(v_1-v_2)=0$) Now, from (1) and (2): $v_1-v_2=0\implies v_1=v_2\implies$ one-one.
<aside> 💡 A function $f:V\rightarrow W$ is bijective(or a bijection) if it is one-one and onto.
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<aside> 💡 A linear transformation $f:V\rightarrow W$ between two vector spaces $V$ and $W$ is said to be an isomorphism if it is a bijection.
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Let $V$ be a vector space with basis $\{v_1,v_2.....,v_n\}$.
Let $f:V\rightarrow W$ be a linear transformation. Then the ordered vectors $f(v_1),f(v_2),...,f(v_n)$ uniquely determine $f$.
Take a General linear transformation $f:\R^n \rightarrow \R^m$, then: