Is it A and C. Because by linear combinations of the vectors in A/C we can get R2.
stutiRastogi
I agree A and C are bases, and can also see that others are not.
But what is the proof to show D is not a basis? Intuitively any linear combination of the vectors in D, would just give R1, since they lie in the same line. Is there a more concrete (algebraic/geometric) way to show this?
nmrrs
A basis for Rn must always consist of exactly n linearly independent vectors. Because the two vectors in D are clearly not linearly independent, we can only construct vectors parallel to them.
stutiRastogi
Yeah that makes sense.
I was able to get it with an example. If we take the vectors to be say u=(1,0) and v=(-1,0), then to get any point (x,y) such that (x,y) = au + bv, where a and b are scalars, we don't get two equations to solve for a and b.
Thanks @nmrrs!
keenan
Right. Also, (D) is not even a basis for the line, since a basis for an $n$-dimensional space must consist of no more and no less than $n$ linearly independent vectors. (For a line, we should have $n=1$, but here there are two vectors.)
amaeda
The cardinality of the basis is called the dimensionality of the vector space, and all bases have the same cardinality. It's useful since it gives the notion of space and how it's constructed by some dimensions.
Animagus
For (D), even we consider it as a 2 dimension basis it's still not working, because it forms no angle with the other dimension which will never represents the volume on the other.
keenan
@Animagus Correct. In other words, even though we have two different vectors, they are not linearly independent; they still span only a 1-dimensional subspace of the plane. So we can't write an arbitrary vector as a linear combination of these two.
Is it A and C. Because by linear combinations of the vectors in A/C we can get R2.
I agree A and C are bases, and can also see that others are not.
But what is the proof to show D is not a basis? Intuitively any linear combination of the vectors in D, would just give R1, since they lie in the same line. Is there a more concrete (algebraic/geometric) way to show this?
A basis for Rn must always consist of exactly n linearly independent vectors. Because the two vectors in D are clearly not linearly independent, we can only construct vectors parallel to them.
Yeah that makes sense.
I was able to get it with an example. If we take the vectors to be say u=(1,0) and v=(-1,0), then to get any point (x,y) such that (x,y) = au + bv, where a and b are scalars, we don't get two equations to solve for a and b.
Thanks @nmrrs!
Right. Also, (D) is not even a basis for the line, since a basis for an $n$-dimensional space must consist of no more and no less than $n$ linearly independent vectors. (For a line, we should have $n=1$, but here there are two vectors.)
The cardinality of the basis is called the dimensionality of the vector space, and all bases have the same cardinality. It's useful since it gives the notion of space and how it's constructed by some dimensions.
For (D), even we consider it as a 2 dimension basis it's still not working, because it forms no angle with the other dimension which will never represents the volume on the other.
@Animagus Correct. In other words, even though we have two different vectors, they are not linearly independent; they still span only a 1-dimensional subspace of the plane. So we can't write an arbitrary vector as a linear combination of these two.