Matrices, Direct Sums and Quotient Spaces — Studies in Algebra and Group Theory

8 Matrices, Direct Sums and Quotient Spaces

19 September 2025 · Denis Auroux

Matrices of linear maps

Last time: given a basis of , we get an isomorphism We represent by a column vector .

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 That is, the columns of give the components of in the basis .

We then have So if is represented by the column vector (i.e. ), then is represented by Think of this as where we treat the product symbolically as matrix multiplication: row column .

Change of basis

What if we choose a different basis for and/or ?

If we change basis from to , write and get an matrix whose th column gives the components of in the basis . Symbolically, .

So , i.e. the element of described by a column vector in the new basis is described by in the old basis.

More conceptually: !

Do the same for , in reverse: let , i.e. . Hence i.e.

In particular, if and we change basis, then for , the matrices and are related by

But... the whole point of linear algebra is to avoid all this and work with linear maps in a coordinate-free language as much as possible.

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 8.1.

Sums and direct sums of subspaces

Returning to the finite case...

Definition 8.2. Given subspaces of some vector space ,

the span of is . Say the span if .

Definition 8.3.

The are independent if with implies for all .
If the are independent and span , say we have a direct sum decomposition . That is, every element of can be written uniquely as with .

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

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

Case of two subspaces:

are independent iff (since 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 ,

are subspaces of and respectively.

is called the rank of .

Theorem 8.4 (Dimension formula).

Proof. Start by choosing a basis for , and complete it to a basis of . We claim is a basis of . Indeed:

If , then write and get , so the span .
If , then , so , i.e. for some . But since are linearly independent, this forces all (and ). Hence the are linearly independent.

So now, since is a basis of , , where (the are a basis of ) and (the are a basis of ). ∎

Corollary 8.5. Given a linear map , there exist bases of and in which the matrix of has the form

Proof. Take the basis of which is as above (the being a basis of ), and complete (a basis of ) to a basis of . ∎

Corollary 8.6. For subspaces ,

Proof. Consider the map from to , Then and , so ∎

Notes not reached in this lecture, covered at the start of {{LINK:L09}}

Quotient spaces

Let be a vector space over a field , a subspace.

Definition 8.7. The quotient space is the space of cosets of in , with addition and scalar multiplication .

The linear map is surjective, with kernel . Hence we get an exact sequence By the dimension formula (), we have

Remark.

Given a linear map , if (i.e. ), then factors through , i.e. there exists such that :

(Define ; this is independent of the choice of in the coset.) Conversely, given , has . Hence
There is a bijection Conversely, (and since ).