MA22020: Advanced linear algebra

Chapter 4 The structure of linear operators

4.1 On normal forms

Question. Given \(\phi \in L(V)\), is there a basis with respect to which \(\phi \) has a “nice” matrix?

Of course, this does not make much sense without some idea of what “nice” should mean for matrices but a reasonable idea might be that there should be a low number of non-zero entries.

There is a matrix version of the same question. For this, recall:

Definition. Matrices \(A,B\in M_n(\F )\) are similar if there is an invertible matrix \(P\in M_n(\F )\) such that
\(\seteqnumber{0}{4.}{0}\)
\begin{equation*} B=P^{-1}AP. \end{equation*}

We can then ask:

Question. Is \(A\) similar to a “nice” matrix?

and a very practical question:

Question (Similarity problem). When are \(A,B\in M_n(\F )\) similar?

A possible answer to this last question would be to compare “nice” matrices similar to \(A\) and \(B\) (recall that similarity is an equivalence relation!).

We already know one situation where this sort of thing works out. Recall from Algebra 1B¹ that \(A\in M_n(\F )\) is diagonalisable if and only if it has an eigenbasis if and only if it is similar to a diagonal matrix

\begin{equation} \label {eq:13} \begin{pmatrix} \lambda _1&&0\\&\ddots &\\0&&\lambda _n \end {pmatrix}. \end{equation}

Here \(\lst \lambda 1n\) are the eigenvalues of \(A\) listed with their multiplicities, that is, each \(\lambda _i\) appears \(\am (\lambda _i)\) times. We say that (4.1) is a normal form of \(A\).

We can conclude, after reordering eigenbases if necessary:

¹ Definition 3.3.1

Theorem. Diagonalisable matrices \(A,B\in M_n(\F )\) are similar if and only if they have the same eigenvalues and multiplicities up to order.

Our plan in this chapter is to try and generalise these ideas to arbitrary \(A\in M_n(\F )\). We encounter two difficulties almost immediately.

(1) Not enough eigenvalues: Let
\(\seteqnumber{0}{4.}{1}\)
\begin{equation*} A= \begin{pmatrix} 0&-1\\1&0 \end {pmatrix}. \end{equation*}

Then \(\Delta _A=x^2+1\) which has no eigenvalues at all in \(\F =\R \). We solve this problem by working over \(\C \).
(2) Not enough eigenvectors: Let
\(\seteqnumber{0}{4.}{1}\)
\begin{equation*} A= \begin{pmatrix} 0&1\\0&0 \end {pmatrix}. \end{equation*}

Then \(\Delta _A=x^2\) but \(\ker A=\Span {(1,0)}\). We therefore do not have enough eigenvectors to span \(\C ^2\). To solve this problem will need a new idea (see §4.3).

In this chapter, we will, among other things, completely solve the similarity problem for any \(A\in M_n(\C )\). This will take quite a bit of work but here is a sneak preview: any \(A\in M_n(\C )\) is similar to a matrix of the form

\begin{equation*} \begin{pmatrix} \lambda _1&*&&0\\ &\ddots &\ddots &&\\ &&\ddots &*\\ \bigzero &&&\lambda _n \end {pmatrix} \end{equation*}

with eigenvalues with multiplicity on the diagonal, each \(*\) on the first super-diagonal either \(0\) or \(1\) and zeros elsewhere.