how to find a matrix relative to a basis

Question

I was wondering how to prove that something is an operator, and that it is linear. I have this formula:

$T[p(x)] = (x − 1)p′(x) + p′′(x)$.

I need to prove that it is an operator and it is linear. I know that to be linear it has to satisfy linearity, and most operators do, but I'm not sure how to apply it to the context of this question?

I was also then wondering how to calculate the matrix of T relative to the standard basis {$1, x, x_2 , x_3$}? does this mean I just insert $1, x, x^2$ in etc.?

Answer 1

HINT.- $$p(x)=ax^3+bx^2+cx+d\\p'(x)=3ax^2+2bx+c\\p''(x)=6ax+2b$$ $$T(p(x))=3ax^3+(2b-3a)x^2+(c-2b+6a)x+2b-c$$

$T$ is a function of $V$ in $V$ where $V=\{$polynomials of degree$\le 3\}$

$$►T(\lambda p(x))=3a\lambda x^3+(2b\lambda-3a\lambda)x^2+(c\lambda-2b\lambda-6a\lambda)x+(2b\lambda+-c\lambda)=\lambda T(p(x))$$

$$►T(p(x)+q(x))=3(a+a_1)x^3+\cdots+(c+c_1-2(b+b_1-6(a+a_1))\\T(p(x)+q(x))=T(p(x))+T(q(x))$$

For the matrix you have at the beginning $$x=0x^3+0x^2+x+0\\x^2=0x^3+x^2+0x+0\\x^3=x^3+0x^2+0x+0\\1=0x^3+0x^2+0x+1$$ Now $$T(x)=x-1\\T(x^2)=\cdots\text {etc }$$

Answer 2

It is linear because$$T\bigl(\alpha p_1+\beta p_2)(x)=\alpha p_1(x)+\beta p_2(x).$$Now, compute $T(1),T(x),T(x^2),T(x^3)$ and expreass each of them as $a_0+a_1x+a_2x^2+a_3x^3$. The matrix that you're after is the matrix whose columns are the coefficients of these vectors.