Help with axler's definition of a polynomial (linear algebra done right)

Question

I'm having trouble following Axler's polynomial definition. I don't understand the meaning of $$p\colon {\bf F}\to {\bf F}$$ on page 10 of Axler

Further, how would I go about rigorously verifying that ${\cal P}({\bf F})$ is a vector space?

Answer 1

The issue/point is that the book defines the set of polys by viewing polys as $\bf F$-valued functions, and then uses the properties of their image $\bf F$ to define the corresponding properties for functions. For instance:

• if $p,q$ are $F$-valued functions, then Axler defines $r = (p+q)$ to be the unique function such that $r(z ) = p(z)+ q(z)$, for all $z\in {\bf F}$.

• Likewise, the scalar multiplication is defined 'pointwise' - namely, if $a\in {\bf F}$, and again viewing a poly $p$ as a function, then we define $s = a p$ to be the unique function such that $s(z) = a p(z)$, for all $z\in {\bf F}$.

• The 'O' function is also defined pointwise, namely by the rule $0(z)= 0$, for all $z$.

• etc.

Now, since Axler gives the definitions 'pointwise', as above, the space of functions automatically inherits all of commutative, associative, distributive, additive inverse - and so on - properties from ${\bf F}$. In particular, the set of ALL functions from ${\bf F}$ to ${\bf F}$ is automatically a vector space, if the vector space operations/axioms are given pointwise. So I would say you don't really have to verify commutativity, say, as you wanted to do in your comment, if you believe/understand/say the 'inherit' argument.

On the other hand, given the Axler's function/pointwise definitions, you do have to verify that the ${\cal P} (F)$ is CLOSED under addition, scalar multiplication, etc. For instance,

• the function $0 \in {\cal P}({\bf F})$, because it agrees for all $z$ with the poly that has coefficients $a_i = 0$, for all $i$ (Axler's notation);

• if $p$ is the function that be written as the poly $p(z) = a_0 +a_1 z+ \cdots + a_n z^n,$ then the poly $-a_0 - a_1 z - \cdots - a_n z^n$, viewed as a function, equals the function $-p$, defined pointwise, as Axler does. Hence ${\cal P}({\bf F})$ is closed under negation (i.e., has additive inverses).

• and so on.

Okay?

Your 'show polys vanishing at $3$ is a subspace' question - I would say that your way of doing it works, because polys that vanish at $3$ all have a factor of $z-3$, as you wrote - and that is something that one can prove. (Don't forget in that approach, that $0 = (z-3) 0$, so that you have the zero element.)

On the other hand, you could do it more 'abstractly' - by 'pure thought', as they say - almost without proving or calculating anything... Namely, the set $V_3$ of ALL functions vanishing at $3$ is a sub vector-space of the set of ALL functions (from $\bf F$ to $\bf F$): for instance,

• if $f(3) =0$, and $g(3)=0$, then $(f+g)(3) = f(3) + g(3) = 0 + 0 = 0$, so $V_3$ is closed under addition, etc.

The abstract fact you would need if $A$ and $B$ are sub vector-spaces of some ambient vector $E$, then the intersection $A\cap B$ is also sub-vector space of $E$. For instance, if

• $x,y\in A\cap B$, then $x+y\in A$ (because $A$ is closed under addition). For the same reason $x+y \in B$. Therefore $x+y \in A\cap B$, and $A\cap B$ is closed under addition.

• and so on...

From that 'abstract' point of view, one would take $A= V_3$, and $B= {\cal P}({\bf F})$, to conclude as a special case that the polys that vanish at $z=3$ is a vector space.

I hope I haven't weirded you out further....

The psychological problem is that we are all first taught to think vectors to be $n$-tuples, i.e., elements of ${\bf F}^n$. Now ${\bf F}^n $ is the arch-typical example of a $n$-dimensional vector space - and yes, every $n$-dimensional $ \bf F$ vector space is isomorphic to (i.e., looks like) ${\bf F}^n$. For example, the vector space of polys of at most degree $4$ is isomorphic to ${\bf F}^5$ (namely, use $a_0,\cdots, a_4$).

On the other hand, usually, the isomorphism is not canonical (i.e., basically, equality) - one needs a choice of coordinates. For instance (only for instance), if somebody standing in the steppes of Siberia envisages the expanse around as a two dimensional vector space, how does one match that vector space 'canonically' with the one that one gets from standing somewhere in Antartica? The two vector-spaces are isomorphic to each other and to ${\mathbb R}^2$, but they are not 'canonically' isomorphic to each other.

Hope this helps - and not the opposite! Good luck...