Recent content by B18

Both of your replies helped me understand what was going on here much better. Thank you.

This is the solution I got. However I am a bit confused on my basis. How can the x4 slot both have zeros in the basis vectors and still span the space? Did I make a mistake along the way? Thanks for the help everyone.

It would have been helpful if our professor explained what pivots were. Thanks though Izzy I'll make sense of what you explained and go from there.

The maximum dimension of the space spanned would have to be m+1, correct? For example if the vectors were from R3 we would need 4 column vectors so that they were linearly dependent.

Homework Statement Let A be an m x n matrix with m<n. Prove that the columns of A are linearly dependent. Homework Equations Its obvious that for the columns to be linearly dependent they must form a determinate that is equal to 0, or if one of the column vectors can be represented by a...

Well I got the following equations from the matrix: x1+x2+x4=0 x3+x5=0 x4=0 Not really sure how I can parametrize these to solve for the basis of the solution space.

Homework Statement Find a basis for and the dimension of the solution space of the homogenous system of equations. 2x1+2x2-x3+x5=0 -x1-x2+2x3-3x4+x5=0 x1+x2-2x3-x5=0 x3+x4+x5=0 Homework EquationsThe Attempt at a Solution I reduced the vector reduced row echelon form. However the second row...

In this case the zero vector would have to be equal to 1?

Sure thing!

Wow, that one flew right past me. Apparently I wasn't thinking about exponent rules enough. The set is all real numbers though, nothing indicating that it is only positives.

Here is the list of axioms I'm using. I never thought of that. Thanks Ray.

Homework Statement Determine if they given set is a vector space using the indicated operations. Homework EquationsThe Attempt at a Solution Set {x: x E R} with operations x(+)y=xy and c(.)x=xc The (.) is the circle dot multiplication sign, and the (+) is the circle plus addition sign. I...

Ok I figured. So everything looks in order here? Just was expecting it to be a little more involved!

Homework Statement Let u, and v be vectors in Rn, and let c be a scalar. c(u+v)=cu+cv The Attempt at a Solution Proof: Let u, v ERn, that is u=(ui)ni=1, and v=(vi)ni=1. Therefore c(ui+vi)ni=1 At this point can I distribute the "c" into the parenthesis? For example: =(cui+cvi)ni=1...