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Determine whether or not the following is a linear transformations.

$L: P_2 \rightarrow P_3, L(p(x)) = 5p(1)+x^2p'(x)$

My main question for this problem is whether the process I used is correct, because I have seen different ways of solving similar questions like this. I used a different process to arrive to the solution.

$\text{If L is a linear transformation mapping a vector space V into a vector space W,} \\ p(x)= a_0+a_1x , \\ q(x) = b_0+b_1x \\ \mathbf{(I)}\qquad L(p(x)+q(x)) = L(p(x))+L(q(x)) \\ = L(a_0+a_1x+b_0+b_1x) \\ 5(a_0+a_1+b_0+b_1) +x^2(a_1+b_1)=5(a_0+a_1+b_0+b_1)+x^2(a_1+b_1) \color{lime}{LHS} \checkmark\\L(p(x)+q(x)) \\ = L(a_0+a_1x)+L(b_0+b_1x) \\ 5(a_0+a_1)+x^2(0+a_1)+5(b_0+b_1)+x^2(0+b_1)\\5(a_0+a_1)+x^2a_1+5(b_0+b_1)+x^2b_1\\=5(a_0+a_1+b_0+b_1)+x^2(a_1+b_1) \color{blue}{RHS}\checkmark \\ \mathbf{(II)} \qquad L(\alpha(p(x)) =\alpha L(p(x)) \\ L(\alpha(a_0+a_1x))\\L(\alpha a_0 +a_1x\alpha) \\ 5(\alpha a_0 +a_1 \alpha)+x^2 a_1 \alpha \color{lime}{LHS} \checkmark \\\alpha L (a_0 +a_1x) = \alpha(5(a_0+a_1)+x^2a_1) \\ 5(a_0\alpha+a_1 \alpha)+x^2 a_1 \alpha \color{red}{RHS} \checkmark \\ \text{Yes Linear Transformation}$

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Yes, it is correct (assuming that $P_2$ is the space of all polynomials whose degree is less than $2$), but you don't have to write each polynomial as $a+bx$. For instance, in order to prove that $L\bigl(p(x)+q(x)\bigr)=L\bigl(p(x)\bigr)+L\bigl(q(x)\bigr)$, all you need to do is\begin{align}L\bigl(p(x)+q(x)\bigr)&=5\bigl(p(1)+q(1)\bigr)+x^2(p(x)+q(x)\bigr)'\\&=5p(1)+5q(1)+x^2\bigl(p'(x)+q'(x)\bigr)\\&=5p(1)+x^2p'(x)+5q(1)+x^2q'(x)\\&=5p(1)+x^2p'(x)+5q(1)+x^2q'(x)\\&=L\bigl(p(x)\bigr)+L\bigl(q(x)\bigr).\end{align}

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    $\begingroup$ I think $P_n$ is the space of all polynomials whose degree is $\leq n$, not $<n$. $\endgroup$ – kccu May 17 '18 at 21:47
  • $\begingroup$ @kccu I suspect so. And my approach also works in that case without any change. $\endgroup$ – José Carlos Santos May 17 '18 at 21:49
  • $\begingroup$ Right. I just wanted to make sure OP knew the correct definition of $P_n$. $\endgroup$ – kccu May 17 '18 at 21:50
  • $\begingroup$ Where did you find the definition for $P_n$ the space of all polynomials? Also is $P_n$ usually < n or $\le$ n? How would you know if the question is < n or $\le$ n $\endgroup$ – Jon May 17 '18 at 22:07
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    $\begingroup$ @Jon You are wrong now. A polynomial with degree less than $2$ is a polynomial of the type $a_0+a_1x$. $\endgroup$ – José Carlos Santos May 17 '18 at 22:19
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Your proof is correct. You can somewhat shorten it by noticing that if the function is linear, then it has a matrix with respect to the standard bases of $P_2$ and $P_3$.

Since $L(1)=5$ and $L(x)=5+x^2$, the matrix should be $$ \begin{bmatrix} 5 & 5 \\ 0 & 0 \\ 0 & 1 \end{bmatrix} $$ Thus the coordinates of $L(a_0+a_1x)$ should be $$ \begin{bmatrix} 5 & 5 \\ 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a_0 \\ a_1 \end{bmatrix}= \begin{bmatrix} 5a_0+5a_1 \\ 0 \\ a_1 \end{bmatrix} $$ corresponding to $$ L(a_0+a_1x)=(5a_0+5a_1)+a_1x^2 $$ which is exactly the action of $L$. The function is linear.

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  • $\begingroup$ If I may have a moment of your time, do I also need to confirm that $L(\alpha p(x) + \beta q(x) = L(\alpha p(x)) + L(\beta q(x))$ $\endgroup$ – Jon May 18 '18 at 22:49
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    $\begingroup$ @Jon No, this already proves it, because it show $L$ is a composition of linear maps. $\endgroup$ – egreg May 19 '18 at 5:58

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