Additional Exercises to 2nd Edition of Axler

The notation used will not be exactly the same as Axler. For instance, we will denote the set {a1,,an}\{a_1, \ldots, a_n\} as {ai}\{a_i\} out of laziness.


Axler does many things right. It jumps straight into the main protagonists of linear algebra — vector spaces — without boring a reader with the details of a field or a vector space. This is all sensible for a first read, but it is useful to eventually learn what a field and vector space actually are.

Field axioms

Definition. A field is a set of elements FF, including two special distinct elements 00 and 11, with two binary operators ++ and \cdot such that

for all elements aa, bb, cFc \in F.

Even though the field is really the triple (F,+,)(F, +, \cdot), we will simply refer to the field as FF for brevity.

Typically we omit \cdot when it is clear, e.g. we will write aba\cdot b as abab.

Examples of fields include \mathbb{Q} and \mathbb{R}, as well as the integers modulo pp, i.e. 𝔽p\mathbb{F}_p. Note that \mathbb{Z} is not a field.

A couple of results you should try to prove:

Vector space axioms

Definition. A vector space consists of a set VV of vectors — including a special vector 00 — over a field FF of scalars, and two binary operators +:V×VV+ \colon V \times V \to V and :F×VV\cdot \colon F\times V \to V which satisfy

for all uu, vVv \in V and aa, bFb \in F.

Note that ++ can either represent the binary operator +:F×FF+ \colon F \times F \to F or +:V×VV+ \colon V \times V \to V. Similarly, \cdot can either represent the binary operator :F×FF\cdot \colon F \times F \to F or :F×VV\cdot \colon F \times V \to V. In the interest of conciseness, we will not explicitly differentiate the two. You will have to infer.

Furthermore, we denote the vector space as VV for brevity, even though it really also involves a field and two binary operators.

A result you should try to prove:


Existence of direct complement for infinite vector spaces

Given a finite-dimensional vector space VV and a subspace UU of VV, there exists some subspace WW of VV such that UW=VU \oplus W = V. (We can prove this by explicitly extending a basis of UU.) However, what if VV is infinite? Then our explicit extension of a basis doesn’t work, because infinite vector spaces cannot have a finite basis.

In fact, the existence of this direct complement is guaranteed not through a proof but through the Axiom of Choice. (This statement being true, in other words, is equivalent to AoC.)

Chapter 2

Chapter 3

Matrix exercises (feel free to skip)

Chapter 5

Chapter 6

Chapter 7


Theorem 7.25 states any normal operator can be expressed as a block diagonal matrix with blocks of size 11 or 22 (and the size 22 block matrices are scalar multiples of the rotation matrix).

This statement isn’t terrible (knowing the explicit representation of a linear map is useful, I guess), but there’s a much more natural way to state it. Return to the Spectral Theorem: (normal/self-adjoint) operators in (\mathbb{C}/\mathbb{R}) have an orthonormal basis of eigenvectors. A (somewhat contrived) reformulation of the Spectral Theorem is that TT can be decomposed into invariant orthogonal subspaces of dim 1\text{dim } 1. And obviously every subspace of dim 1\text{dim } 1 is self-adjoint (and thus normal), so we can say

TT can be decomposed into normal invariant orthogonal subspaces of dimension 11 iff it is (normal/self-adjoint) in (\mathbb{C}/\mathbb{R}).

So the equivalent reformulation of 7.25 would be

TT can be decomposed into normal invariant orthogonal subspaces of dimension 11 or 22 iff it is normal in \mathbb{R}.

Chapter 8


Let’s restate 8.5 and 8.9 in their full forms, which Axler alludes to later in the chapter.

  • (Higher-powered 8.5) There exists some non-negative integer mm such that null Tknull Tk+1\text{null } T^k \subsetneq \text{null } T^{k+1} for k<mk < m and null Tk=null Tk+1\text{null } T^k = \text{null } T^{k+1} for kmk \geq m.
  • (Higher-powered 8.9) There exists some non-negative integer mm such that range Tkrange Tk+1\text{range } T^k \supsetneq \text{range } T^{k+1} for k<mk < m and range Tk=range Tk+1\text{range } T^k = \text{range } T^{k+1} for kmk \geq m.