S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 2: Rings, Fields and Vector Spaces

7 Bases, Dimension, and Matrices of Linear Maps

· Denis Auroux

Original Professor Auroux’s handwritten notes (PDF), behind a password for now:

Span, linear independence, basis

Let be a vector space over .

Definition 7.1 (Span). Given , the span of is the smallest subspace of which contains . Concretely, Say span if .

Definition 7.2 (Linear independence). We say are linearly independent if

Definition 7.3 (Basis). are a basis of if they are linearly independent and span .

Then any element of can be expressed uniquely as for some .

Equivalently, given , we have a linear map and

  • are linearly independent is injective;
  • span is surjective;
  • are a basis is bijective.

Example 7.4. and are a basis of . So are and for most fields . (What if ?)

Note. We will see soon: if has a basis with elements, then every basis of has elements. We say the dimension of is .

More generally, let be a vector space over and a subset (not necessarily finite).

Definition 7.5 (Span, independence, basis for arbitrary subsets).

  • = the smallest subspace of containing (all finite linear combinations of elements of ).
  • The elements of are linearly independent if there are no finite linear relations: for all , all , and all distinct elements of ,
  • is a basis of if its elements are linearly independent and span .

Example 7.6.

  • is a basis of .
  • Does have a basis? What is it?

Linear maps

Definition 7.7 (Linear map). Let be vector spaces over . A homomorphism of vector spaces, or linear map, , is any map that is compatible with the operations:

Proposition 7.8. The set of linear maps is itself a vector space over , denoted .

Proof. Given and , define One can check that

  • and defined in this way are linear maps (so we do have operations on );
  • these operations on satisfy the axioms of a vector space.

(Rather boring, but worth checking if you’re not sure!) ∎

We’ll soon see: if and , then . (In bases for and , linear maps become matrices!)

The notions depend on the field

The notions of basis, dimension, linearity do depend on the coefficient field !

Example 7.9. is a vector space over itself (of dimension 1; is a basis). It is also a vector space over (of dimension 2, with basis ).

If are -vector spaces, hence also -vector spaces, any -linear map is also -linear, but the converse isn’t true: .

For example, complex conjugation is -linear: and for all ; but it is not -linear ().

Bases and dimension

Say is finite-dimensional if there exists a finite set that spans .

Lemma 7.10. If spans , then a subset of is a basis.

Proof. If the are linearly independent, they form a basis. Otherwise, there is some linear relation , with the not all zero. This can be solved for = a linear combination of the others if . So remove : still spans . Continue removing elements until the remaining ones are linearly independent. ∎

Thus, every finite-dimensional vector space has a basis.

Lemma 7.11. If are linearly independent, there exists a basis of which contains .

Proof. Let be a spanning set for . By induction we enlarge to a basis of for each .

For : is a basis of .

Assume is a basis of . If , then we already have a basis of . Otherwise, are linearly independent (why?) and span .

This ends with a basis of (since span). ∎

Theorem 7.12. If and are bases of , then (same number of elements).

Proof.

  • We claim there exists such that is a basis.

    Indeed, are linearly independent, but don’t span (else gives a linear relation ). So there exists such that (else can’t span all of ).

    Now are linearly independent (why?), but using all the ’s, we can write (necessarily ). So and this implies span , hence are a basis.

  • Repeat this process to exchange one for one each time. (We don’t use the same twice, since the new we pick has to be independent of the rest of our basis.)

We end up with only ’s, and get an -element subset of that is also a basis. Necessarily this is all of , and . ∎

Definition 7.13 (Dimension). The dimension of is the cardinality of any basis.

Given a basis of , we get a linear map Linear independence injective, and spanning surjective, so is an isomorphism!

Every finite-dimensional vector space over is for , and a basis gives a specific choice of such an isomorphism. We represent by a column vector .

Notes not reached in this lecture, covered at the start of {{LINK:L08}}

Matrices of linear maps

Given bases of and of , we can represent a linear map by an matrix . This amounts to:

Write with given by multiplication with column vectors: .

Notation: , the matrix of in the given bases.

The entries of are characterized by i.e. the columns of give the components of in the basis .

Representing any element as the column vector , and similarly for , .

To be very explicit:

As a memory aid, the isomorphism given by the basis can be written symbolically as multiplication of row and column vectors: (⚠ the aren’t numbers!!), and then (compare above).

This construction gives an isomorphism between the vector spaces and ! (Linear maps matrices.) In particular .

Direct sums and products of vector spaces

Given vector spaces and , (with componentwise operations).

Similarly, given vector spaces, .

But for an infinite collection , we have two different constructions:

Example 7.14.

Sums and direct sums of subspaces

Returning to the finite case…

Definition 7.15 (Sum, independence, direct sum). Given subspaces of some vector space :

  • the span of is . Say the span if ;
  • the are independent if , for all ;
  • if the are independent and span , say we have a direct sum decomposition , i.e. every element of can be written uniquely as with .

Relation to the previous notion: for each we have an inclusion map . These assemble into a linear map span is surjective; they are independent is injective. If both hold, then is an isomorphism and we have a direct sum decomposition.

In this case (get a basis of by taking the union of bases of ).

Proposition 7.16 (Case of two subspaces).

  • are independent iff . (Because iff .)
  • . (We’ll see this soon.)
  • iff and .

Also note: given a subspace , there exists another subspace such that . ( is definitely not unique!) To find : take a basis of , complete it to a basis of , and let .

Rank and the dimension formula

Given finite-dimensional vector spaces and and a linear map ,

  • and

are subspaces of and .

Definition 7.17 (Rank). is called the rank of .

Theorem 7.18.

Proof: next time.