help understanding basis and dimension

I am working on the following two problems:

Consider the subset U = {(x1,4x1-7x2,x2) : x1, x2 are real numbers} of R^3

Assume that U is a subspace of R^3.

Find a basis for the subspace U = {(x1,4x1-7x2,x2) : x1, x2 are real numbers} of R^3 and explain why it is a basis for U.

I have deduced that U has the correct number of elements which is 3.

I need to show that U spans R^3 and that U is linearly independant.

However I am for some reason having a hard time in doing so... I'm pretty sure this is all I need to do? But just can't get this problem started.


Part b) asks to find the dimension of said subspace. The dimension (according to my notes) of R^n with standard operations is n. So in this case it is 3 right ?

U is the image of a linear map from R2 to R3 given by $\displaystyle f(x_1,x_2)=(x_1,4x_1-7x_2,x_2)$. It's pretty easy to see that f is zero only at (0,0), (i.e. f has trivial kernel), so that the image (U) has the same dimension as the domain (R2). Whenever that happens, you can take a basis of the domain, and the image of the basis set forms a basis of the image. So {f(1,0), f(0,1)} is a basis for U.

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Originally Posted by battleman13

I am working on the following two problems:

Consider the subset U = {(x1,4x1-7x2,x2) : x1, x2 are real numbers} of R^3

Assume that U is a subspace of R^3.

Find a basis for the subspace U = {(x1,4x1-7x2,x2) : x1, x2 are real numbers} of R^3 and explain why it is a basis for U.

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Any vector in that subspace is of the form $\displaystyle (x_1, 4x_1- 7x_2, x_2)= (x_1, 4x_1, 0)+ (0, -7x_2, x_2)= x_1(1, 4, 0)+ x_2(0, -7, 1)$.

That tells you the basis and dimension immediately.

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[I have deduced that U has the correct number of elements which is 3.

I need to show that U spans R^3 and that U is linearly independant.

However I am for some reason having a hard time in doing so... I'm pretty sure this is all I need to do? But just can't get this problem started.


Part b) asks to find the dimension of said subspace. The dimension (according to my notes) of R^n with standard operations is n. So in this case it is 3 right ?

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No, it isn't! The dimension of $\displaystyle R^3$ is 3 but this is a subspace of $\displaystyle R^3$. The fact that every vector in the subspace depended upon two numbers, $\displaystyle x_1$, and $\displaystyle x_2$ should hve been a clue!