Hi,
problem:
Consider the vector space and the subsets consisting of those vectors (polynomials) for which
(a) x has degree 3,
(b) 2x(0)=x(1)
(c) whenever ,
(d) for all .
In which of these cases is a vector space?
attempt:
I first assume that is a vector space over as this is not mentioned in the problem.
(a) Not a vector space.
Closed under vector addition.
Closed under scalar multiplication.
It does not have an origin that is a polynomial of degree 3 such that .
(b) Is a vector space.
Closed under vector addition.
Closed under scalar multiplication.
There exists an element (origin) , such that
(c) and (d), could need a push in the right direction here.
Thanks!
No, it is not closed under vector addition.
If and then which is NOT third degree.
No, it is not closed under scalar multiplication. 0 is a scalar and 0 time any third degree polynomial is NOT a third degree polynomial.Closed under scalar multiplication.
It does not have an origin that is a polynomial of degree 3 such that .Thanks!But of course, that just says this is NOT a vector space- but you could have stopped after showing it is not closed under addition.
You need to show these, not just say they are true. If 2x(0)= x(1), 2y(0)= y(1) and z(t)= x(t)+ y(t), then 2z(0)= 2(x(0)+ y(0))= 2x(0)+ 2y(0)= x(1)+ y(1)= z(1) and, for any scalar, a, 2(az(t))= 2(ax(t)+ ay(t))= a(2x(t))+ a(2y(t))= ax(1)+ ay(1)= a(x(1)+ y(1))= az(1).(b) Is a vector space.
Closed under vector addition.
Closed under scalar multiplication.
Good! You showed that the 0 function is in this set.There exists an element (origin) , such that
Suppose x(t) is such that for [/tex]0\le t\le 1[/tex]. What about (-1)x?(c) and (d), could need a push in the right direction here.
(d) is pretty much like the others. If x(t)= x(1-t), y(t)= y(1-t), and z= x+ y, then z(t)= x(t)+ y(t)= x(1-t)+ y(1-t)= z(1- t), etc.
(c)
, i.e. not closed under scalar multiplication.
(d)
Scalar multiplication:
Origin:
Now I guess I need to show that .
Again, thank you very much for dumbing down replies for me HallsOfIvy
I dream of the day you won't have to!