S. D. V. Stephens

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

10 Eigenvectors and Upper-Triangular Matrices

· Denis Auroux

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

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 .

  • 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.

Eigenvectors and eigenvalues

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

Definition 10.1. 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 10.2. . 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 HW4, Problem 1).

Proposition 10.3. Eigenvectors of with distinct eigenvalues are linearly independent.

Proof. Assume are eigenvectors with , the all distinct. Assume there is a linear relation with the not all zero, and that this has the fewest () possible nonzero of any such relation. Then another linear relation! .

Fix such that , and subtract ( times the first relation): This is a linear relation where the coefficient of is now zero, but all other nonzero coefficients (at least one) remain nonzero (since ). This contradicts the minimality assumption. ∎

Corollary 10.4. The number of distinct eigenvalues of is at most , and if equality holds then is diagonalizable.

Algebraically closed fields

Definition 10.5. A field is algebraically closed if every nonconstant polynomial has a root in , i.e. there exists such that .

If so, then by the division algorithm for polynomials, we can write , and repeating, we get

  • Fundamental theorem of algebra: is algebraically closed. (The proof is not pure algebra; we’ll discuss it in Math 55b.)
  • If is not algebraically closed, then there exists an algebraically closed field , constructed from by adjoining roots of polynomials in . E.g. , whereas (fact: polynomials in have roots in ).

Theorem 10.6. If is algebraically closed and a finite-dimensional vector space over , then any linear operator has an eigenvector, i.e. there exist and such that .

Proof. Let , and take any nonzero vector . Then the vectors must be linearly dependent. So there exist (not all zero) such that Since is algebraically closed, we can factor the polynomial , hence (!! the product here is composition of operators, but this is legit !!).

Now has a nontrivial kernel (), which implies that at least one of the is not an isomorphism; hence there exist and such that , i.e. . ∎

Notes not reached in this lecture, covered at the start of Lecture 11

Upper-triangular form

Corollary 10.7. Given over an algebraically closed field , there exists a basis of in which the matrix of is upper-triangular, (i.e. each subspace is invariant).

Proof. Induction on .

  • If , then any nonzero vector gets mapped to a multiple of itself ✓ (any matrix is triangular).

  • Assume the result is true for dimension , and consider with . Since is algebraically closed, has at least one eigenvector , . Let , and let , with the quotient map (i.e. work “mod ”). Since , , so factors through . So there exists such that commutes (intuitively, “ mod ”).

    Since , by the induction hypothesis there exists a basis of such that . Let be such that (i.e. ). (Note: span span , hence are a basis.) Then (In fact, in the basis ,

∎

Now suppose we have and a basis of such that is upper-triangular, i.e. each is an invariant subspace of . Denote by the diagonal entries of .

Lemma 10.8. is invertible iff all the diagonal entries of are nonzero.

Proof.

  • If all are nonzero, then is surjective (hence an isomorphism) since etc. for all .
  • If , then , so has nontrivial kernel (since ); hence , and is not invertible.

∎

Corollary 10.9. The following are equivalent:

  1. is an eigenvalue of ;
  2. is not invertible;
  3. for some diagonal entry of any upper-triangular matrix representing .

((1) (2) since eigenvectors , and (2) (3) by applying the lemma to and the matrix .)

Next goal: further study of invariant subspaces and eigenvalues for linear operators over an algebraically closed , especially : generalized eigenspaces and Jordan normal form. (This is Axler ch. 8; we’ll return to the skipped chapters 6 and 7 soon.)