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.
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...
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.
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...
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...