Find a basis of the given span{[2,1,0,-1],[-1,1,1,1],[2,7,4,5]}
So I got the RREF, and found the basis to be two rows of the RREF, which are [1,0,-1/3,0] and [0,1,2/3,1], but the answer is [2,1,0,-1],[-1,1,1,1]
Where did I do wrong?
Let T: V\rightarrowW be a linear transformation.
1. T is said to be onto if I am T=W.
2. T is said to be one-to-one if T (v)=T(v_{1}) implies v=v_{1}.
I think I understand the definitions, it just seems so hard to apply them into questions:frown:
I thought I understood what u meant:redface:
For my question, say a function A: P→P maps a polynomial f(x) to the polynomial f(x). Then
A(x)=x^{2}
A(x+2)=(x+2)^{2}
So this function is onto, but not one-to-one?:rolleyes:
Hmmm, let f: P-->P be the mapping that takes x to x^{2}, this function is onto but not one-to-one? (Is this correct for this question?)
I still cannot think of a function that is one-to-one, but not onto:confused:
Say function T:A-->B, if onto but not one-to-one, then every element from B from get mapped to from A. If one-to-one but not onto, then some or every element from B get mapped to from A? But how can I come up with the function then?:confused:
So if it is onto, the linear combination should have infinitely many solutions?
If it is one-to-one, then the linear combination should have only one solution?
write P for the vector space of all polynomials, a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}, , a_{0}, a_{1},...,a_{n}\inR, n=0,1,2...
1. Find a linear transformation P->P that is onto but not one-to one
2. Find such a linear transformation, that is one-to-one but not onto
I have been thinking...
Hi, Zondrina,
Thanks for your input.
The original question was "Prove that the set V=R^2 with addition defined by (x,y)+(x'+y')=(x+x'+1, y+y') and scaler multiplication by k(x,y)=(kx+k-1, ky) is a vector space, find-(x+y) and the zero vector in this vector space."
I guess the question was...
Clearly, that was a typo with my first post, "((-1)x+(-1)-1), (-1)y)=(-x-1, -y)", so the result should be (-x-2, -y), not (-x-1, -y)
so I guess I was on the right track.
When you mention " Denoting the additive identity, for now, as (a, b), we must have (x, y)+ (a, b)= (x, y) for all (x...