Hey n22.
Hint: What are the sub-space axioms? (There are three of them).
Hi I need help with this .
The lecturers at uni always say that I must test if zero is in the vector space .if its not ,then apparently its not a vector space.
But sometimes its not always possible.-when is it not possible?
THanks.
This is the actual question.
Let X={f which is a element of F such that f(x)=A+Be^x, A,B is an element of R}
Justify and determine if X is subspace of F.
Hi Chiro,
1.that S is not empty (check by chucking in zero)
2.S closed under addition
3.S is closed under scalar multiplication
f(o)=A+B...its not equal to zero so....does that matter?when does it mater and when does it not?
Let g(x)=C+De^x ....
by addition,A+C +(B+D)e^x
kf(x)=KA+kBe^x..
i feel like the set up is kindda wrong..
Do you mean f(0)? That is not relevant. You want the zero vector, that is the function such that f(x)= 0 for all x, in the set.
A and B here can be any real numbers. What do you get if you take A= B= 0?
Which is of the form "A+Be^x" with A+ C instead of A and B+ D instead of B.Let g(x)=C+De^x ....
by addition,A+C +(B+D)e^x
You probably don't need to specifically state that A+ C and B+D are real numbers!
Which is of the form "A+ Be^x" with kA instead of A and kB instead of B.kf(x)=KA+kBe^x..
i feel like the set up is kindda wrong..