Basis and Dimension

Basis and Dimension
Let V be a vector space, and let B={v1,…,vn} be a set of vectors in V. B is a basis for V if:
- B is linearly independent
- B is a generating set of V (V is the span of the vectors in B)
In other words, a basis B is the smallest set such that every vector in V is a linear combination of the vectors in B.
Example 1
B=⎩⎨⎧100,010,001⎭⎬⎫ is a basis for R3 because:
- The vectors in B are linearly independent
- Every vector in R3 can be generated from the vectors in B
Note: removing any vector would mean that the set is no longer a generating set, so it is as small as possible.
Example 2
B={[1000],[0010],[0100],[0001]} is a basis for M2×2.
Dimension
The dimension of a vector space Vis the number of vectors in its basis, which we denote dim(V).
Example 1
B=⎩⎨⎧100,010,001⎭⎬⎫ is a basis for R3, so dim(R3)=3.
Example 2
B={[1000],[0010],[0100],[0001]} is a basis for M2×2, so dim(M2×2)=4.

Example: Verifying a Basis
Let V be the vector space {(x,y)∣x,y∈R+} with operations:
- (x1,y1)+(x2,y2)=(x1x2, y1y2)
- k⋅(x,y)=(xk,yk)
Show that B={(10,1),(1,10)} a basis for V. [Hint: 0=(1,1)]
Step 1
Show that the vectors in B are linearly independent.
c1⋅(10,1)+c2⋅(1,10)=0⟹⟹⟹⟹(10c1,1c1)+(1c2,10c2)(10c1,1)+(1,10c2)(10c1⋅1,1⋅10c2)(10c1,10c2)====(1,1)(1,1)(1,1)(1,1)
So we need 10c1=1 and 10c2=1.
The only solution to these equations is c1=c2=0, so B is linearly independent.
Step 2
Show that B is a generating set for V.
We must show that every vector (x,y) in V can be written as a combination of the vectors in B:
(x,y)=====c1⋅(10,1)+c2⋅(1,10)(10c1,1c1)+(1c2,10c2)(10c1,1)+(1,10c2)(10c1⋅1,1⋅10c2)(10c1,10c2)
From this we can see that:
x=10c1y=10c2⟹⟹c1=log10(x)c2=log10(y)
For every pair (x,y)∈V: (x,y)=log(x)⋅(10,1)+log(y)⋅(1,10)
Note that x,y∈R+ so the logarithms are well-defined.
Since we can find appropriate coefficients for any values of x and y, B is a generating set for V.
Therefore, B is a basis of V.

Example: Finding a Basis
Let V be the vector space {(x,y)∣x,y∈R+} with operations:
- (x1,y1)+(x2,y2)=(x1x2, y1y2)
- k⋅(x,y)=(xk,yk)
Consider the following subspace of V: S={(x,y)∈V∣x3y=1}.
Find a basis for S, and find the dimension of S.
[Hint: for any real number x>0 and any real number k: xk=eln(xk)=(ek)ln(x)]
Step 1
To find a basis, first find a generating set for S.
If x=(x,y)∈S is an arbitrary vector, then we know that: x3y=1⟹y=x31⟹y=x−3
So x=(x,y)=(x,x−3).
Goal: Write x as a linear combination of vectors in S.
We will "factor out" the variable x using the defined operations of V along with the hint:
(x,x−3)====(x1,x−3)((e1)ln(x),(e−3)ln(x))ln(x)⋅(e1,e−3)ln(x)⋅(e,e−3)using the hintby the definition of scalar multiplication
Note that (e,e−3)∈S because x3(e)3y(e−3)=1 (or simply: y=x−3).
We can generate any vector in S as a scalar multiple of (e,e−3): (x,y)=(x,x−3)=ln(x)⋅(e,e−3)
Therefore, the generating set is B={(e,e−3)}.
Step 2
Bis linearly independent because it contains only one vector (and it is not the zero vector).
Therefore, B is a basis for S. Since B contains only one vector, S is a 1-dimensional subspace of V.