S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 3: Linear Maps and Operators

9 Dual Spaces and Linear Operators

· Denis Auroux

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

Quotient spaces

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

Definition 9.1. 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.

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

    Commutative triangle: the map phi from V to W equals the quotient map q from V to V/U followed by the induced map phi-bar from V/U to W.

    (Define ; this is independent of the choice of in the coset.) Conversely, given , has . Hence

  2. There is a bijection Conversely, (and since ).

Dual spaces

Let be a vector space over a field .

Definition 9.2. The dual vector space is the space of linear functionals on , i.e. linear maps :

Example 9.3. If , any tuple , , determines a map Conversely, let be the standard basis of . Given , let ; then i.e. . So

More generally, given a finite-dimensional and a basis , any linear map is determined by the values , so we get an isomorphism Equivalently, we get a basis of consisting of the linear functionals such that and for . (Then !) This is called the dual basis!

However, there is not a natural map , despite the above about bases. Each element of the dual basis depends not just on , but on all the ’s. There’s no such thing as “the dual of a vector”.

On the other hand, we do have a natural map (“evaluation”)

If is finite-dimensional, then by working in the basis , the dual basis , and the double dual basis , we see that ✓, and so ; hence is an isomorphism. Thus:

Proposition 9.4. If is finite-dimensional, then is an isomorphism.

When is infinite-dimensional, is injective, but not an isomorphism! The reason is: assume has a basis , so every element of is uniquely with only finitely many nonzero (). Then for all , is a well-defined element of , characterized by for all .

So , which is larger, and the linear functionals (, for all ) do not span . (We can complete them to a basis via Zorn’s lemma.) A similar enlargement happens again when passing from to .

Annihilators

Definition 9.5. The annihilator of a subspace is (This is a subspace of .)

  • The restriction map , , is surjective with kernel , so we have an exact sequence This in turn implies .
  • Also, we’ve seen above that . Hence
  • Either way, this implies

The transpose of a linear map

Definition 9.6. Given a linear map , the transpose, or dual, of , , is defined as follows: given a linear functional , composing with gives a linear map . Thus (Check: is linear.)

Check:

  • Given a basis of , elements of are represented by column vectors , and elements of are represented by row vectors . Applying a linear functional to a vector corresponds to .
  • If , then This is because: given and , , so , viewed as an operation on row vectors, is . Meanwhile the dual bases give a description of elements of , by column vectors, which are the transposes of the row vectors. The claim then follows since as a column vector is .

Proposition 9.7. (In the finite-dimensional case)

  • is injective iff is surjective;
  • is surjective iff is injective.

This follows from:

Proposition 9.8.

  1. .
  2. , assuming finite dimension.

Proof.

  1. .
  2. If , then , so . So . The dimension formula and (1) imply , hence the inclusion is an equality.

∎

Notes not reached in this lecture, covered at the start of {{LINK:L10}}

Linear operators

A linear operator on (aka endomorphism of ) is a linear map .

Notation: .

  • When using a basis to express by a (square) matrix, we want to use the same basis on each side: transforms by .

  • New thing: we can compose linear operators with each other, , or with themselves, , or even apply polynomials: So is a (noncommutative) ring.

  • Given vector spaces and linear operators , we can define an operator on . The operator leaves the subspaces invariant: ; and working in a basis of such that , , the matrix of is block diagonal: Conversely, if and , then is of this form.

    More generally, if we only assume and is invariant () but not necessarily , then the matrix of would be block triangular:

So: a typical way to study is to look for invariant subspaces.

  • If is invariant and (so for some ), then necessarily for some .

Definition 9.9. An eigenvector of is a vector , , such that for some . is called the eigenvalue corresponding to .

  • If we can find a basis of consisting of eigenvectors of , then we have diagonalized , i.e. found a basis where its matrix is diagonal: This is the best outcome, but not always possible!

Example 9.10. . The matrix has eigenvectors and (or any multiples) with eigenvalues and .

However, has only one eigenvector (up to scaling!), with eigenvalue : it is not diagonalizable.

doesn’t even have any eigenvectors (see homework).