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Basis (linear algebra)

Redirected from Basis of a vector space A subset S of a vector space V is said to be a basis of V if it satisfies one of the four equivalent conditions:

1. S is a minimal generating set of V.
2. S is a maximal set of linearly independent vectors.
3. S is both a set of linearly independent vectors and a generating set of V.
4. every vector in V can be expressed as a linear combination of vectors in S in a unique way.
Recall that a set S is a generating set of V if every vector in V is a linear combination of vectors in S.

This definition tacitly includes a finiteness condition. A basis of a vector space is sometimes called a Hamel basis in order to distinguish it from the concept of an orthonormal basis of a Hilbert space and some other kinds of bases that occur in Banach spaces. An orthonormal basis of a Hilbert space H is an orthonormal set of members of H such that any member of the H can be written as a linear combination of a possibly infinite set of members of the orthonormal basis. In order to speak of infinite linear combinations, one must have a notion of convergence. Since Hilbert spaces are metric spaces, such a notion is available and natural.

Using Zorn's lemma, one can show that:

every vector space has a basis.
Every basis of a vector space has the same cardinality, called the dimension of the vector space.

Example I: Show that the vectors (1,1) and (-1,2) form a basis for R2.

Proof: We have to prove that these 2 vectors are both linearly independent and that they generate R2.

Part I: To prove that they are linearly independent, suppose that there are numbers a,b such that:

• a(1,1)+b(-1,2)=(0,0)
Then:
• (a-b,a+2b)=(0,0) and a-b=0 and a+2b=0.
Subtracting the first equation from the second, we obtain:
• 3b=0 so b=0.
And from the first equation then:
• a=0.

Part II: To prove that these two vectors generate R2, we have to let (a,b) be an arbitrary element of R2, and show that there exist numbers x,y such that:

• x(1,1)+y(-1,2)=(a,b)
Then we have to solve the equations:
• x-y=a
• x+2y=b.
Subtracting the first equation from the second, we get:
• 3y=b-a, and then
• y=(b-a)/3, and finally
• x=y+a=((b-a)/3)+a.

Example II: We have already shown that E1, E2, ..., En are linearly independent and generate Rn. Therefore, they form a basis for Rn.

Example III: Let W be the vector space generated by et, e2t. We have already shown they are linearly independent. Then they form a basis for W.