Mathematics 49 Online
OpenStudy (anonymous):

Does anyone have a good grasp on the ideas of linear algebra? E.g. The geometric nature of linear equations.

OpenStudy (anonymous):

?

OpenStudy (anonymous):

why don't you post your problem and we'll find out?

OpenStudy (anonymous):

For which real numbers x do the vectors: (x, 1, 1, 1), (1, x, 1, 1), (1, 1, x, 1), (1, 1, 1, x) not form a basis of R 4 ? For each of the values of x that you ﬁnd, what is the dimension of the subspace of R 4 that they span

OpenStudy (anonymous):

By observation, x = 1 would make them all the same vector, and their span would only be one dimensional. So that is one value of x that wont let them span. The systematic way of find the values would be to make those vectors into columns and put them in a matrix, then row reduce the matrix to reduced echelon form.

OpenStudy (anonymous):

first if x = 1 you've got a 1 dimension subspace

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

another solution by observation is x = -3

OpenStudy (anonymous):

if you had the matrix (call it A): x 1 1 1 1 x 1 1 1 1 x 1 1 1 1 x and x = -3, then a solution to the equation: $A\vec u = 0$ would be u = (1, 1, 1, 1), which means the Null Space of A has something in it other than the 0 vector, which means the Null Space has at least dimension 1, which means the columns space cant be dimension 4, which means the columns dont span R^4 (lol)

OpenStudy (anonymous):

your problem is the liberty of (1,1,1,x), (1,1,X,1), (1,x,1,1) and (x,1,1,1) because this family is always generating R4 in order to find the case where it is not a base, try to prove that your family is free, then you ll find the case where it doesn't go well.

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