1. ## Linear Algebra Problem

Hi,

I was trying this problem and I got stumped. Could someone please explain this problem to me? Thank you!

Let T: P3->R be given by t(p(x)=p''(-4)
A) Compute T(2x^3+5x^2-4x+3)
B) show that T is in fact a linear transformation
C) Show that Ker(T) = P1
D) Show that Range(T) = R
E) Show that T can be written as a composition.

2. Part A) :
Notice that $p(x)=2x^3+5x^2-4x+3$. So $p'(x)=6x^2+10x-4$. So $p''(x)=12x+10$. Now evaluate this polynomial in $x=-4$ and you have the answer.
B) What's the definition of a linear transformation?
Edit : C) We want to know for what polynomials in $P_3$ the relation $T(p(x))=0$ holds. For it, we chose the general form of polynomials that are in $P_3$. $p(x)=ax^3+bx^2+cx+d$.
We know that $T(p(x))=p''(-4)$, so $T(ax^3+bx^2+cx+d)=6a(-4)+2b=-24a+2b$.
So $T(ax^3+bx^2+cx+d)=0 \Leftrightarrow -24a+2b=0 \Leftrightarrow b=12a$. Clearly $P_1 \in \ker (T)$, however $P_1 \neq \ker (T)$. Are you sure the question reads
Show that Ker(T) = P1
?

3. According to the book, T(K*U)=kT(U). How would I apply that here? Multiply K for the inside of the function? it does not seem to make sense to me.

4. Originally Posted by egshih
According to the book, T(K*U)=kT(U). How would I apply that here? Multiply K for the inside of the function? it does not seem to make sense to me.
Well as stated here, it's not sufficient. $T(ku+v)=kT(u)+T(v)$ must holds.
Yes you have to multiply " $u$" by a scalar, say $k$. $u$ is a polynomial and $v$ is another one. So for example take $u=ax^3+bx^2+cx+d$ and $v=ex^3+fx^2+gx+h$.
You have to show that $T(ku+v)=kT(u)+T(v)$ holds.

Please notice that I edited my above post.

5. Thank you for all the help with this problem. I was looking through the problem and I think there is a mistake on this problem. In my notes, it says that the T(p(x)) should be equal to P''(x) for this problem to be correct. Also off topic, how do you post the math equations as an image?

6. Originally Posted by egshih
Thank you for all the help with this problem. I was looking through the problem and I think there is a mistake on this problem. In my notes, it says that the T(p(x)) should be equal to P''(x) for this problem to be correct. Also off topic, how do you post the math equations as an image?
You're welcome.
As of Latex (math images), check out the subforum http://www.mathhelpforum.com/math-help/latex-help/.

7. Originally Posted by egshih
Hi,

I was trying this problem and I got stumped. Could someone please explain this problem to me? Thank you!

Let T: P3->R be given by t(p(x))=p''(-4)
i'll skip (A) since it is pretty straight forward, and Arbolis dealt with it, you just have to stop before plugging in -4

B) show that T is in fact a linear transformation
for this you must show that

(a) T(u + v) = T(u) + T(v) and (b) T(ku) = kT(u) .....where u,v are elements of P3 and k is a scalar.

these properties actually follow directly from the properties of derivatives

(a) T(u + v) = (u + v)'' = u'' + v'' = T(u) + T(v) ....the second equal sign follows from the "distributive" (not sure if that's what it's called) property of differentiation

(b) T(ku) = (ku)'' = ku'' = kT(u) .........again, the second equal sign follows by the property of derivatives that says we can factor out a constant multiplier of a function when differentiating

C) Show that Ker(T) = P1
this is simply saying that the set of elements of P3 that map to zero under this transformation is precisely the first degree polynomials. clearly any first degree polynomial will go to zero when differentiated twice...

D) Show that Range(T) = R
here you want to show that T maps onto R. this is true, since for all r in R, the polynomial p3(x) = (r/2)*x^2 + cx + d will map to it.

E) Show that T can be written as a composition.
T can be written as D $\circ$D , where D is the differential operator

EDIT: i just saw post #5 so i made some changes. without throwing that -4 in the mix, things should work out nicer.