Are you really clear on what a basis is?

Any vector is of the form <x, y, z>= <2t, t, 4t>= t<2, 1, 4>. Can you answer the question now?the line: x = 2t, y = t, z = 4t

The equation is the same as y= xthe plane x - y = 0

Any vector is of the form <x, y, z>= <x x, z>= <x, x, 0>+ <0, 0, z>= x<1, 1, 0>+ z<0, 0, 1>

all vectors of the form (x; y; z), where y = x + z

Try this one yourself now.

I understand the basis but I cant seem to understand these??

Q2) WI guess it would but then it would be even easier to replace the entire problem with "1+ 1"! You aree have stated that [1; x; x^2; ......; x^n] is a basis for Wn, where Wn denotes the vector space of polynomials of maximal degree n. Verify that [1; x -1; x^2 - x; : : : ; x^n -x^n-1] is also a basis for Wn

Now for this i was generally thinking of subbing in values for the relevant x's but is this correct?

Q3)Let V be a vector space with basis [v1; v2; v3]. Which of the following

collections are also a basis for V ?

1) [v1, v1 + v2, v1 + v3]

2) [2v1,3v2, v1 + v2];

3) [v1, v2 + v3, v1 -v3, v1 + v2]

4) [v2, v3 - v, v3 + v2]

5) [ v2, v3- v1]

6) [v1 - v2 + 2v3 , 2v2 + v3, 3v1 + v2 -3v3]

Again is it easier to just make values for v1,v2,v3 up ?givenv1, v2, and v3. You are told that v1, v2, v3 is a basis. Any basis for an n dimensional vector space has three properties: the vectors span space, the vectors are independent, and there are n vectors. And anytwoof those implies the third.

Here, all of the given collections have three vectors so it is sufficient to show that each v1, v2, v3 can be written as a linear combination of the given vectors.