 # For the transformation L:C^1[a,b]->C[a,b] definied by L(f)=dfdx (means, derivative of function f w.r.t x) Jaxson Mack 2022-08-07 Answered
For the transformation $L:{C}^{1}\left[a,b\right]\to$ $C\left[a,b\right]$ definied by $L\left(f\right)=\frac{df}{dx}$ (means, derivative of function $f$ w.r.t $x$),
a) Show that $L$ is linear.
b) Find $Ker\left(L\right)$.
c) Find $nullity\left(L\right)$.
d) Is $L$ one-to-one? Explain.
e) is $L$ invertible? Explain.
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a) Consider, $c$ be scalar and $f,g\in {c}^{1}\left[a,b\right]$
$L\left(cf+g\right)=\frac{d}{dx}\left(cf+g\right)$
$=\frac{d}{dx}\left(cf\right)+\frac{d}{dx}\left(g\right)$
$=c\frac{d}{dx}\left(f\right)+\frac{d}{dx}\left(g\right)$
$L\left(cf+g\right)=cL\left(f\right)+L\left(g\right)$
$\therefore L$ is linear
b) $Ker\left(L\right)=\left\{f\in {c}^{1}\left[a,b\right]:\frac{df}{dx}=0\right\}$
$Ker\left(L\right)=\left\{\text{constant terms}\right\}$
Basis for $Ker\left(L\right)=\left\{1\right\}$
c) $\therefore Nullity\left(L\right)=1$
d) As we know $L$ is $1-1$ iff $Ker\left(L\right)=\left\{0\right\}$
\therefore $L$ is not $1-1$ $\left(\because Ker\left(L\right)\ne \left\{0\right\}\right)$
c) Any transformation is invertible if it is both $1-1$ and onto. As $L$ is not $1-1$
$\therefore L$ is not invertible
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