Question

why isn't these two sets not basis for the indicated vector spaces:

1) p_1=1+x+x^2, p_2=x-1 for P_2

I said because it is in P_3 and by theorem if r>n then it is linearly dependent and therefore not a basis ( just wanted a verification on my way to think about it)

2) u=(-1,3,2) u_2=(6,1,1) for R^3

I can't seem to figure out but I said it's because it can't form a square matrix ( so I can't find the determinant to decided whether or not it spans R^3)

Answer 1

The dimension of the vector space is equal to the cardinality of the basis.

Answer 2

Both vector spaces have dimension 3, but the proposed basis only has two elements.

Answer 3

http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Answer 4

I see so it can't span it. And I see what you mean r> n would stand for the amount of vectors in the set not the number of components

Answer 5

thanks! haven't covered that in class. weird.

Answer 6

The dimension of a vector space is defined as the cardinality of a basis for the vector space.

Answer 7

It is well defined since every basis for a vector space can be proven to have the same number of elements.

Answer 8

Did you talk about dimension yet in class?

Answer 9

briefly, very very briefly and it was after we covered basis. (redoing problems for my final and the book doesn't cover dimensions really until the set right after basis) I just assumed R^3 was in 3rd dimension. and so on

Answer 10

but then, can't i just prove it is either 1) linearly depedent and / or 2) that it doesnt span the given space?

Answer 11

If you want to avoid using dimension, you can show that you can find another linearly independent vector to add to the set.

Answer 12

u_1=(-1,3,2) u_2=(6,1,1) are linearly independent but do not span \(\mathbb{R}^3\) therefore it is not a basis. To prove this you should try to find a vector that is not in the span.

Answer 13

ah I see how it can't span R^3. But I was taught to prove that a set is either spanning or not by finding its determinant, and finding the determinant of a 2x3 is not possible

Answer 14

and how would I go about to find a vector not in the span? the cross product? it should be an orthogonal line to the plane create by the two vectors no?

Answer 15

The cross product will work

Answer 16

But it only helps you in a 3 dimensional space.

Answer 17

so in lower / higher dimensions Im guessing I need to solve for a vector such that k_1*v+k_2*u+k_3*w= i and change one of the variable of the vector that is a combination of the three ( that would be for R^4)

Answer 18

one of the components* sorry

Answer 19

I still think in the end the best way to answer the question is to talk about dimension. It is trivial if you do.

Just refer to the dimension theorem in your answer: http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Answer 20

If in class you discussed using matrices, another way is to place the two vectors as columns in a matrix.

Answer 21

Let the third column be all 0s

Answer 22

You will notice that the determinant is of course 0.

Answer 23

What about the fact that one of the vectors is not in the set it is spanning? Does this matter?

Answer 24

A basis for a vector space is a set of linearly independent vectors that span the vector space. Therefore, if you have some set of linearly independent vectors \(\mathcal{B}\), that set is not a basis if you can find another vector in the vector space not in the span of \(\mathcal{B}\).