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3.2 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 \(L(V,V)\)).
A special case of the analysis of §1.4.2 tells us that linear operators in the presence of a basis are closely related to square matrices: if \(V\) is a finite-dimensional vector space over \(\F \) with basis \(\cB =\lst
{v}1n\) and \(\phi \in L(V)\) then the matrix of \(\phi \) with respect to \(\cB \) is the square matrix \(A=(A_{ij})\in M_n(\F )\) with
\(\seteqnumber{0}{3.}{0}\)
\begin{equation}
\label {eq:12} \phi (v_j)=\sum _{i=1}^nA_{ij}v_i,
\end{equation}
for \(\bw {1}jn\).
Equivalently, \(\phi (\lc {x}v1n)=\lc {y}v1n\) where
\(\seteqnumber{0}{3.}{1}\)
\begin{equation*}
\by =A\bx .
\end{equation*}
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:
\(\seteqnumber{0}{3.}{1}\)
\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
\(\seteqnumber{0}{3.}{1}\)
\begin{equation}
\label {eq:28} \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:
\(\seteqnumber{0}{3.}{2}\)
\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*}
This construction plays nicely with the algebra of polynomials:
-
Proposition 3.4. For \(p,q\in \F [x]\), \(\phi \in L(V)\) and \(A\in M_n(\F )\),
\(\seteqnumber{0}{3.}{2}\)
\begin{align}
\label {eq:29} (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:30}.
\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
\(\seteqnumber{0}{3.}{4}\)
\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 (3.3) for \(\phi \).
Now for (3.4). We have
\(\seteqnumber{0}{3.}{4}\)
\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 (3.2) 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
\(\seteqnumber{0}{3.}{4}\)
\begin{equation*}
pq(\phi )=qp(\phi )=q(\phi )p(\phi )
\end{equation*}
by what we have already proved. □