- #1

gruba

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## Homework Statement

Check if [itex]L(p)(x)=(1+4x)p(x)+(x-x^2)p'(x)-(x^2+x^3)p''(x)[/itex] is a linear transformation on [itex]\mathbb{R_2}[x][/itex]. If [itex]L(p)(x)[/itex] is a linear transformation, find it's matrix in standard basis and check if [itex]L(p)(x)[/itex] is invertible. If [itex]L(p)(x)[/itex] is invertible, find the function rule of it's inverse.

## Homework Equations

-Linear transformations

## The Attempt at a Solution

[itex]L(p)(x)[/itex] is a linear transformation if

[itex]L(z(x)+w(x))=L(z)(x)+L(w)(x) \forall z,w\in\mathbb{R_2}[x][/itex] and

[itex]L(\alpha z(x))=\alpha L(z)(x)\forall z\in \mathbb{R_2}[x],\forall\alpha\in\mathbb{R}[/itex]

Let [itex]p(x)=a+bx+cx^2\Rightarrow L(p)(x)=a+(4a+2b)x+(3b+c)x^2[/itex].

Let [itex]z(x)=p+qx+rx^2[/itex] and [itex]w(x)=\alpha+\beta x+\gamma x^2[/itex].

[itex]L(z(x))=p+(4p+2q)+(3q+r)x^2[/itex] and [itex]L(w(x))=\alpha+(4\alpha+2\beta)+(3\beta+\gamma)x^2[/itex] then [itex]L(z(x))+L(w(x))=(p+\alpha)+(4p+2q+4\alpha+2\beta)x+(3q+r+3\beta+\gamma)x^2[/itex]

[itex]L(z(x)+w(x))=L((p+\alpha)+(q+\beta)x+(r+\gamma)x^2)[/itex].

From this [itex]\Rightarrow L(z(x)+w(x))=L(z)(x)+L(w)(x)[/itex].

[itex]L(\alpha z(x))=L(\alpha p+\alpha qx+\alpha rx^2)=\alpha p+(4\alpha p+2\alpha q)x+(3\alpha q+\alpha r)x^2[/itex].

[itex]\alpha L(z)(x)=\alpha(p+(4p+2q)x+(3q+r)x^2)[/itex].

From this [itex]\Rightarrow L(\alpha z(x))=\alpha L(z)(x)[/itex] and [itex]L(p)(x)[/itex] is a linear transformation.

Standard basis for [itex]\mathbb{R_2}[x][/itex] is [itex]\mathcal{B}=\{1,x,x^2\}[/itex].

[itex][L]_{\mathcal{B}}=

\begin{bmatrix}

1 & 0 & 0 \\

4 & 2 & 0 \\

0 & 3 & 1 \\

\end{bmatrix}

[/itex].

Is the matrix [itex][L]_{\mathcal{B}}[/itex] correct?

If not, how to construct it?

[itex]L(p)(x)[/itex] is invertible because [itex]\det [L]_{\mathcal{B}}\neq 0[/itex].

[itex][L]^{-1}_{\mathcal{B}}=

\begin{bmatrix}

1 & 0 & 0 \\

-2 & 1/2 & 0 \\

6 & -3/2 & 1 \\

\end{bmatrix}

[/itex]

Function rule of [itex]L^{-1}(p)(x)[/itex] can be read from [itex][L]^{-1}_{\mathcal{B}}\Rightarrow L^{-1}(p)(x)=a+(-2a+\frac{1}{2}b)x+(6a-\frac{3}{2}b+c)x^2[/itex].

Is the function rule of [itex]L^{-1}(p)(x)[/itex] correct?

If not, how to construct it?

Also, how to check if [itex]L(p)(x)[/itex] is bijective?