I was reading the chapter Polynomial from the book Linear Algebra by Hoffman and Kunze,
In the book, polynomial $F[x]$ was defined as a subspace of $F^{\infty}$ (spanned by $x$, $x^2, …$ with $x^k = (0,…,1,0,…$) the $1$ in the k-th entry). And they also defined the multiplication as in normal polynomial functions:
$$(fg)_{i}=\sum_{j=0}^{i} (f)_{j}(g)_{i-j}$$
where $f=(f_{1},f_{2},…)$ and $g=(g_{1},g_{2},…)$, $f,g \in F[x]$.
It was also shown in the book that this implies that $fg$ = $gf$. Now, they also defined:
Also, the theorems:
But now this means that $g(\alpha)f(\alpha)=(gf)(\alpha)=(fg)(\alpha)=f(\alpha)g(\alpha)$. Since the linear algebra is not necessarily commutative, we cannot always have that as true right?
For example, if we take the linear algebra as the space of n x n matrix over $R^{1}$, which is a linear algebra with identity but not commutative $(n\geq 2)$. And we also define $f = x $ and $g = x + 2$, we know generally it is not true that $f(A)g(A) = g(A)f(A)$ since matrix multiplication is not commutative, but this property was implied by the definitions.
Am I missing something? would you please kindly point it out? Thank you !
Best Answer
You're correct that the algebra need not be commutative, but we don't need the full strength of commutativity to know that $f(\alpha)$ and $g(\alpha)$ commute. Consider, indeed, your example of $f = x + 2$, $g = x$. Then $f(A) g(A) = (A + 2) A = A^2 + 2A$. On the other hand, $g(A) f(A) = A (A + 2) = A^2 + A \cdot 2$. However, $A \cdot 2 = 2A$, so $f(A) g(A) = g(A) f(A)$. Here, we used the fact that $A$ commuted with itself and with scalars in $F$. Indeed, these two facts are enough to explain in general why $f(\alpha)$ and $g(\alpha)$ commute. Both are sums of scalar minutes multiples of powers of $\alpha$. Powers of $\alpha$ commute with each other and with scalar multiples, so $f(\alpha)$ and $g(\alpha)$ will commute.