3.2 Linear operators, matrices and polynomials

3.2.1 Linear operators and matrices

  • Definition. Let \(V\) be a vector space over \(\F \). A linear operator on \(V\) is a linear map \(\phi :V\to V\).

    The vector space of linear operators on \(V\) is denoted \(L(V)\) (instead of \(L(V,V)\)).

  • Notation. Write \(M_n(\F )\) for \(M_{n\times n}(\F )\).

Recall from Algebra 1B §1.5 that, in the presence of a basis, there is a close relationship between linear operators and square matrices:

  • Definition. Let \(V\) be a finite-dimensional vector space over \(\F \) with basis \(\cB :\lst {v}1n\). Let \(\phi \in L(V)\). The matrix of \(\phi \) with respect to \(\cB \) is the matrix \(A=(A_{ij})\in M_{n}(\F )\) defined by:

    \begin{equation} \label {eq:27} \phi (v_j)=\sum _{i=1}^nA_{ij}v_{i}, \end{equation}

    for all \(\bw 1jn\).

Thus the recipe for computing \(A\) is: expand \(\phi (v_j)\) in terms of \(\lst {v}1n\) to get the \(j\)-th column of \(A\).

Equivalently, \(\phi (\lc {x}v1n)=\lc {y}v1n\) where

\begin{equation*} \by =A\bx . \end{equation*}

The map \(\phi \mapsto A\) is a linear isomorphism \(L(V)\cong M_{n}(\F )\) which also plays well with composition and matrix multiplication: if \(\psi \in L(V)\) has matrix \(B\) with respect to \(\cB \) then \(\psi \circ \phi \) has matrix \(BA\) with respect to \(\cB \). This gives us a compelling dictionary between linear maps and matrices.

  • Remark. There is a fancy way to say all this: recall that a basis \(\cB :\lst {v}1n\) of \(V\) gives rise to a linear isomorphism \(\phi _{\cB }:\F ^n\to V\) via

    \begin{equation} \label {eq:3} \phi _{\cB }\vec \lambda 1n=\sum _{i=1}^n\lambda _iv_i. \end{equation}

    Now the relation between \(\phi \) and \(A\) is that

    \begin{equation*} \phi =\phi _{\cB }\circ \phi _A\circ \phi _{\cB }^{-1} \end{equation*}

    or, equivalently, \(\phi _{\cB }\circ \phi _A=\phi \circ \phi _{\cB }\) so that the following diagram commutes:

    (math image)

    (The assertion that such a diagram commutes is simply that the two maps one builds by following the arrows in two different ways coincide. However, the diagram also helps us keep track of where the various maps go!)

3.2.2 Polynomials in linear operators and matrices

A special feature of \(L(V)\) is that composition is a binary operation \((\phi ,\psi )\mapsto \phi \circ \psi :L(V)\times L(V)\to L(V)\). Thus we can think of composition as a multiplication of operators which suggests the following notations:

  • Notation. For \(\phi ,\psi \in L(V)\) write \(\phi \psi \) for \(\phi \circ \psi \in L(V)\).

    Similarly, write \(\phi ^{n}\) for the \(n\)-fold composition of \(\phi \) with itself:

    \begin{equation*} \phi ^n=\phi \circ \dots \circ \phi , \end{equation*}

    with \(\phi \) repeated \(n\) times on the right, and define \(\phi ^0:=\id _V\), \(\phi ^1:=\phi \).

    Finally, for \(A\in M_n(\F )\), set \(A^0=I_{n}\), \(A^1=A\).

With these notations and conventions, we have

\begin{equation} \label {eq:7} \phi ^{n+m}=\phi ^n\phi ^m,\qquad A^{n+m}=A^nA^m, \end{equation}

for any \(\phi \in L(V)\), \(A\in M_n(\F )\) and \(n,m\in \N \).

Note that if \(\phi \) has matrix \(A\) with respect to a basis \(\cB \) then \(\phi ^n\) has matrix \(A^n\) with respect to \(\cB \), for all \(n\in \N \).

We can now evaluate polynomials on operators and matrices:

  • Definition. Let \(p\in \F [x]\), \(p=a_0+\dots +a_nx^n\), \(\phi \in L(V)\) and \(A\in M_n(\F )\). Then \(p(\phi )\in L(V)\) and \(p(A)\in M_n(\F )\) are given by:

    \begin{align*} p(\phi )&:= a_0\id _V+a_1\phi +\dots +a_n\phi ^n=\sum _{k\in \N }a_{k}\phi ^k,\\ p(A)&:= a_0I_{n}+a_1A+\dots +a_nA^n=\sum _{k\in \N }a_{k}A^k. \end{align*}

  • Remark. If \(\phi \) has matrix \(A\) with respect to a basis \(\cB \) then \(p(\phi )\) has matrix \(p(A)\) with respect to \(\cB \).

This construction plays nicely with the algebra of polynomials:

  • Proposition 3.4. For \(p,q\in \F [x]\), \(\phi \in L(V)\) and \(A\in M_n(\F )\),

    \begin{align} \label {eq:8} (p+q)(\phi )&=p(\phi )+q(\phi )&(p+q)(A)&=p(A)+q(A)\\ (pq)(\phi )&=p(\phi )q(\phi )=q(\phi )p(\phi )&(pq)(A)&=p(A)q(A)=q(A)p(A)\label {eq:9}. \end{align}

  • Proof. We prove the formulae for \(\phi \). The arguments for \(A\) are very similar.

    Write \(p=\sum _{k\in \N }a_{k}x^{k}\) and \(q=\sum _{k\in \N }b_kx^k\). Then

    \begin{equation*} (p+q)(\phi )=\sum _{k\in \N }(a_k+b_k)\phi ^k=\sum _{k\in \N }a_k\phi ^k+\sum _{k\in \N }b_k\phi ^k=p(\phi )+q(\phi ) \end{equation*}

    which establishes (3.4) for \(\phi \).

    Now for (3.5). We have

    \begin{align*} (pq)(\phi )&=\sum _{k\in \N }\bigl (\sum _{i+j=k}a_ib_j\bigr )\phi ^k=\sum _{k\in \N }\bigl (\sum _{i+j=k}a_ib_j\phi ^i\phi ^j\bigr )\\ &=\sum _{k\in \N }\sum _{i+j=k}(a_i\phi ^i)(b_j\phi ^j) =\bigl (\sum _{i\in \N }a_i\phi ^i\bigr )\bigl (\sum _{j\in \N }b_j\phi ^j\bigr )=p(\phi )q(\phi ). \end{align*} Here we used (3.3) for the last equality on the first line and linearity of \(\phi ^{i}\) to get \(b_j\phi ^i\phi ^{j}=\phi ^i(b_j\phi ^j)\).

    Finally \(pq=qp\) so that

    \begin{equation*} pq(\phi )=qp(\phi )=q(\phi )p(\phi ) \end{equation*}

    by we have already proved.  □

  • Remark. The fancy way to say Proposition 3.4 is that the maps \(p\mapsto p(\phi ):\F [x]\to L(V)\) and \(p\mapsto p(A):\F [x]\to M_n(\F )\) are homomorphisms of rings (see MA22017).