suppose p(x) is in ker(T). what does this mean? it means T(p(x)) = 0 (the polynomial, not the number).
since T(p(x)) = x^{2}p(x), we have 2 possibilities:
a)x^{2} is the 0-polynomial. is this ever true?
b) p(x) is the 0-polynomial. what does this say about dim(ker(T))?
note that dim(P_{2}) = 3 (the "2" is misleading). what does the rank-nullity theorem tell you about dim(im(T))?
a) x^2 would equal zero only if x=0
b) if p(x) is the zero polynomial then wouldn't that mean that dim(ker(T)) = 0
i don't understand what you mean by im(T)
The R_N says dim(range) + dim(kernel) = dim(domain)
I hope I am saying anything right
I am sorry if i am saying something really stupid and it might make me look like an idiot, but I am just trying to read this lesson now and do these questions on the study guild.
so please excuse anything out of this world i might say
im(T) is the image, or range, of T. it's just an abbreviation (im(T) looks cooler than rang(T)).
your conclusion for (a) is incorrect. do not confuse the expression x^{2}, where x is a real number, with the POLYNOMIAL x^{2} (which is a FUNCTION, not a number).
oh yeah I knew that's where i probably went wrong, so then for (a) x^2 can't ever be 0 polynomial
so then are we only dealing with part (b) here when it comes to the kernel so far it seams the only thing in the kernel is the {0}, so its dim is 0.
so that would leave the range to be dim(domain) -dim(kernel). so its 3
well when I was given a polynomial like say (3x^2+2x+5) I would look at it in terms of R^3 (3,2,5) and from there i would work with questions regarding polynomials so i never really had to know more then that till now i guess
can you please elaborate on what i should know, I know I am taking alot of your time, and i really appreciate your efforts so far
ok, that's good. now if you're identifying 3x^{2}+2x+5 with (3,2,5) what is your basis?
that is you have (fill in the blanks): 3___ + 2____ + 5____ ? the stuff in the blanks are your basis vectors.
oh so my x^2, x,1 are my bases vectors.
i see
so to go back to what we were doing you state " perhaps {x2,x3,x4} might form a basis?" if its for P4 or in R^5 (0,0,1,1,1)
would that be enough to form a bases, since what i know is that a bases would have to span and isn't it to small to span all of R^5 or P4
and i think i lost though of how we are relating this to the original question about Kernal and Range, since i still don't know the answer
things you should know (if you don't, re-read your texts)
1. a subset of a linearly independent set is linearly independent
corollary: a subset of a basis is linearly independent
2. the span of a linearly independent set is a subspace for which the LI set forms a basis
corollary: a subset of a basis determines a subspace
P_{4} and R^{5} have the same dimension so they "act alike" (as vector spaces). but they are not "the same" there's just a CORRESPONDENCE between the two:
ax^{4}+bx^{3}+cx^{2}+dx+e <--> (a,b,c,d,e)
inducing the following correspondence between the two bases:
x^{4} <--> (1,0,0,0,0)
x^{3} <--> (0,1,0,0,0)
x^{2} <--> (0,0,1,0,0)
x <--> (0,0,0,1,0)
1 <--> (0,0,0,0,1)
*********
in terms of your linear mapping for your problem, the range of T is {f(x) in P_{4}: f(x) = ax^{4}+bx^{3}+cx^{2}}
that is f(x) = x^{2}(p(x)) , where p(x) = ax^{2}+bx+c.
if you want to think of this as the subspace of R^{5} spanned by (1,0,0,0,0) ("x^{4}"), (0,1,0,0,0) ("x^{3}") and (0,0,1,0,0) ("x^{2}") that's "sort of" ok, just realize you're using an EQUIVALENCE, not an EQUALITY.