Recent content by greendays

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?

I tried to answer it in post 13, as I mentioned, I know the definitions, but have no clue how to use them.

yes, but back to the original question, how do you approach that question? I mean I know the definitions, but how to apply them to that question?

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:

no, I don't think I got the idea, so confused:cry:

f(x)=2x+1 is 1-1 but not onto?

Hi vela, Can you give me hints on this question? I have been thinking about it over and over again, but still can not figure it out: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:

Thanks very much for your input, vela. But how did you get A(1)=2?

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...