a) Show that $L$ is linear.

b) Find $Ker(L)$.

c) Find $nullity(L)$.

d) Is $L$ one-to-one? Explain.

e) is $L$ invertible? Explain.

Jaxson Mack
2022-08-07
Answered

For the transformation $L:{C}^{1}[a,b]\to $ $C[a,b]$ definied by $L(f)=\frac{df}{dx}$ (means, derivative of function $f$ w.r.t $x$),

a) Show that $L$ is linear.

b) Find $Ker(L)$.

c) Find $nullity(L)$.

d) Is $L$ one-to-one? Explain.

e) is $L$ invertible? Explain.

a) Show that $L$ is linear.

b) Find $Ker(L)$.

c) Find $nullity(L)$.

d) Is $L$ one-to-one? Explain.

e) is $L$ invertible? Explain.

You can still ask an expert for help

Siena Bennett

Answered 2022-08-08
Author has **17** answers

a) Consider, $c$ be scalar and $f,g\in {c}^{1}[a,b]$

$L(cf+g)=\frac{d}{dx}(cf+g)$

$=\frac{d}{dx}(cf)+\frac{d}{dx}(g)$

$=c\frac{d}{dx}(f)+\frac{d}{dx}(g)$

$L(cf+g)=cL(f)+L(g)$

$\therefore L$ is linear

b) $Ker(L)=\{f\in {c}^{1}[a,b]:\frac{df}{dx}=0\}$

$Ker(L)=\{\text{constant terms}\}$

Basis for $Ker(L)=\{1\}$

c) $\therefore Nullity(L)=1$

d) As we know $L$ is $1-1$ iff $Ker(L)=\{0\}$

\therefore $L$ is not $1-1$ $(\because Ker(L)\ne \{0\})$

c) Any transformation is invertible if it is both $1-1$ and onto. As $L$ is not $1-1$

$\therefore L$ is not invertible

$L(cf+g)=\frac{d}{dx}(cf+g)$

$=\frac{d}{dx}(cf)+\frac{d}{dx}(g)$

$=c\frac{d}{dx}(f)+\frac{d}{dx}(g)$

$L(cf+g)=cL(f)+L(g)$

$\therefore L$ is linear

b) $Ker(L)=\{f\in {c}^{1}[a,b]:\frac{df}{dx}=0\}$

$Ker(L)=\{\text{constant terms}\}$

Basis for $Ker(L)=\{1\}$

c) $\therefore Nullity(L)=1$

d) As we know $L$ is $1-1$ iff $Ker(L)=\{0\}$

\therefore $L$ is not $1-1$ $(\because Ker(L)\ne \{0\})$

c) Any transformation is invertible if it is both $1-1$ and onto. As $L$ is not $1-1$

$\therefore L$ is not invertible

brasocas6

Answered 2022-08-09
Author has **3** answers

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