Recent content by ykaire

Um, I'm guessing that vectors span R2 and 3 Vectors span R3. so... that must mean that 6 vectors span R6.

I'm not sure, to be honest.

1. Can a set of 5 vectors in R6 span all of R6?I want to say that it does span, because i remember my teacher saying "don't think of the physical world," but I'm not entirely sure if it does.

it reduces to x= u1 + u2

I guess I could write the equation as x= u1 + u2 but, like i said, i don't even know if i even started the problem off correctly.

1. Let A∈ Μ4,3 (that is, a 4 x 3 matrix). Let v1, v2 ∈ R^4 and let w = v1 +v2. Suppose there exists u1, u2 as an element of R^3 such that v1 = Au1 and v2= Au2 prove the Ax=w is consistent Homework Equations 3.i honestly don't know if I'm doing this correctly at all, but this is what I have done...

the minimum number of vectors would be 4? and there are 4 (?) vectors in my spanning set, i believe? so, does that mean that R4 is indeed a span {a1, a2, a3, a4}?

yeah, i see. now, i am also asked if R^4 = span {a1, a2, a3, a4} what is the difference between this question and the last one?

now, i am also asked if R^4 = span {a1, a2, a3, a4} what is the difference between this question and the last one?

oh, ok thank you so much :)

but I'm not exactly sure what to do with it? do i have to solve the other parameters in terms of the free parameter?

but I'm not exactly sure what to do with it? do i have to solve the other parameters in terms of the free parameter?

I can tell that there will be one free variable.

hi! thank you :) ok, I just reduced the matrix and i got: 1 0 2 1 0 1 1 0 0 0 1 -4 0 0 0 0 This means that it is consistent, so a solution does exist. So, would that mean that the span would also exist in R4?

[b]1. i am given a matrix A= 1 0 2 1 1 1 3 1 2 3 8 -2 -3 3 -5 1 and then it asks why do I know that span {a1, a2, a3, a4} are a subset of R^4. Homework Equations The Attempt at a Solution Is it as easy as saying "because there 4 vectors?"