# Verifying linear transformations

P= vector space that consists of polynomials of degree <=2. X:P->P is defined as X(f(x))= f(x) - f'(x)

Verify that X is a linear transformation .

Can anyone help with this.

• I suppose $f'$ means the first derivative?
– user368131
Mar 9 '17 at 21:36
• Yes. @Tortuga.A
– ZDD
Mar 9 '17 at 21:38

Ok, if $f'$ is the first derivative of $f$. Then in order to check if $X$ is linear, we have to check the following 2 things: (I omitted $(x)$ just for simplicity)

1. If $X(f+g)=X(f)+X(g)$? This is easy to check since by the definition of $X$, we have \begin{align} X(f+g)&=(f+g)+(f'+g')\\ &=(f+f')+(g+g')\quad (by\ re-arranging)\\ &=X(f)+X(g) \end{align} So our first argument holds based on our calculation
2. Then we need to show if $X(cf)=cX(f)$ for some scalar $c$? The method is very similar to the proof of part 1, and I will leave it to you.

• X(cf)=(cf) - (cf') = c(f - f') = cX(f) @Tortuga.A
– ZDD
Mar 9 '17 at 21:47
• bingo! But to be more precise, we actually have $X(cf)=(cf) - (cf)'$, which is $X(cf)=(cf) - c(f')$ since derivative is linear. Anyway you are correct.
– user368131
Mar 9 '17 at 21:49
• Thank You! Although the question is solved, I faced a problem with another question but would like to apply to it. If C be consisting of the basis of polynomials 1,x, and x^2. Then compute the matrix [X]C How would this be solved. @Tortuga.A
– ZDD
Mar 9 '17 at 21:55
• You know what $X$ does to a polynomials $f$, so you can find what $X(1), X(x),X(x^2)$ are, and then you can write them as a linear combination of the basis $\{1,x,x^2\}$, and then you will get your matrix.
– user368131
Mar 9 '17 at 21:59
• So X(1)=1, X(x)=x+1, X(x^2)=x^2 +2x How would write this as a combination?
– ZDD
Mar 9 '17 at 22:04

$P$ is a vector space over $\mathbb{R}$, so, to prove that the transformation $T(f(x))=f(x)+f'(x)$ is linear you have to prove that:

$\forall \alpha \in \mathbb{R}$ and $\forall f, g \in P$ we have $T(f+\alpha g)=T(f)+\alpha T(g)$

To do this write the transformation and use the linearity of derivative: $$T(f+\alpha g)=(f+\alpha g)+(f+\alpha g)'=f+ \alpha g +f'+\alpha g'= f+\alpha f' + g+\alpha g' =T(f)+\alpha T(g)$$