Suppose there was a basis for and a certain number of dimensions for subspace W in R4.Why is the number of dimensions 2? W={⟨4s−t,s,t,s⟩∣s,t∈R} For instance, apparently, {⟨0,1,4,1⟩,⟨1,1,3,1⟩} is a valid set, and it happens to be of dimension 2 in R4. Does a basis for Rn have to have n vectors?

Question

Suppose there was a basis for and a certain number of dimensions for subspace W in R4.Why is the number of dimensions 2?  W={⟨4s−t,s,t,s⟩∣s,t∈R}  For instance, apparently, {⟨0,1,4,1⟩,⟨1,1,3,1⟩}  is a valid set, and it happens to be of dimension 2 in R4. Does a basis for Rn have to have n vectors?

nebajcioz · Accepted Answer

The 3rd and the 4th coordinates are the only independent ones. The first two can be expressed in terms of the last two.

terorimaox · Accepted Answer

The dimension of a subspace is decided by its bases, and not by the dimension of any vector space it is a subspace of.  The dimension of a vector space is defined by the number of vectors in a basis of that space (for infinite dimensional spaces, it is defined by the cardinality of a basis). Note that this definition is consistent as we can prove that any basis of a vector space will have the same number of vectors as any other basis.  In the case of Rn we know that dim⁡(Rn)=n as {(1,0,0,…,0),(0,1,0,…,0),…(0,0,…,0,1)} is a basis for Rn and has n elements.  In the case of W={⟨4s−t,s,t,s⟩∣s,t∈R} we can write any element in W assu→+tv→ where u→=(4,1,0,1) and v→=(−1,0,1,0).  From this, we have that {u→,v→} is a spanning set for W. Because u→ and v→ are clearly not scalar multiples of each other (note the positions of the 0s) that means that {u→,v→} is a linearly independent spanning set for W, that is, a basis. Because W has a basis with 2 elements, we say that dim⁡(W)=2.  Note that the dimension of a vector space is not dependent on the whether its vectors may exist in other vector spaces of larger dimension. The only relation is that if W is a subspace of V then dim⁡(W)≤dim⁡(V) and dim⁡(W)=dim⁡(V)⇔W=V

Suppose there was a basis for and a certain number of dimensions for subspace W in RR^4.Why is

Answered question

Answer & Explanation

New Questions in Linear algebra