Recent content by Izzy

That they're 0. So the only way to make the second sum 0 is if ##c_2## is also 0. And since the second sum is equivalent to the first sum, all the coefficients must still be 0, so w, ##v_2, ..., v_n## are linearly independent? Sorry Ray Vickson! Thought you were joking, not confused.

So, do you mean that I should just say ##c_1(v_1 + cv_2) + c_2 v_2 + \cdots + c_n v_n = 0## ##c_1v_1 + (c_1c+c_2)v_2 + ... + c_nv_n = 0 ## is linearly independent because ##v_1, v_2, ... v_n## are linearly independent? Also, I know HallsofIvy said it, I just didn't go back and quote him. I...

I think I got it. ##0 = c_1 w + c_2 v_2 + \cdots + c_n v_n = 0## Let ##c_3, ... c_n = 0##, and now we'll show that ##c_1w + c_2v_2## = 0. ##c_1(v_1+cv_2) + c_2v_2 = 0## ##c_1v_1+c_1cv_2 + c_2v_2 = 0## ##c_1v_1 + (c_1c+c_2)v_2 = 0## ##c_1## must be 0, because it's the only way to get rid of...

I tried it like that first, but I couldn't actually figure out how to show that ##\vec{w}, \vec{v}_2, \ldots, \vec{v}_n## are linearly independent. Can I have a hint for how to do that? Also, I'm a girl btw. :P

Show that the reduced row echelon form of the mxn matrix will have at most m pivots. Then there are n-m columns without pivots, which can all be expressed as linear combinations of the columns with pivots. Since they are linear combinations of other columns, they are linearly dependent.

Homework Statement Let V be a vector space, and suppose that \vec{v_1}, \vec{v_2}, ... \vec{v_n} is a basis of V. Let c\in\mathbb R be a scalar, and define \vec{w} = \vec{v_1} + c\vec{v_2}. Prove that \vec{w}, \vec{v_2}, ... , \vec{v_n} is also a basis of V. Homework Equations If two of the...