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Our book states that there is an isomorphism from $Hom(V,W)$, the vector space of all linear transformations from $V$ to $W$, to the matrix space $M_{m\times n}^F$ which is defined by $T:\rightarrow [T]^B_C$ where $B$ and $C$ are bases of $V$ and $W$ respectively.

There are actually two questions here which I think are related.

  1. It's not entirely clear to me what the nature of the object in $Hom(V,W)$ are. Are they the explicit formulas for $T$ according to the given basis? For example T(x,y)=(2x+y,x+y) according to the given basis?

  2. This is related to another question I asked. Can the explicit formula for a linear transformation be according to anything other than the standard basis?

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The matrices in $M_{m\times n}$ are explicit formulas in terms of bases. The objects in $Hom$ are abstract maps without reference to bases :) –  rschwieb Jun 15 '12 at 11:58

3 Answers 3

$\,(1)\,\,\,$The objects in $\,\hom(V,W)\,$ are abstract linear transformations, without any need of specify basis, though one can use them to give a little more concrete form to the maps.

$\,(2)\,\,\,$If by "explicit formula" you mean something like what you wrote with coordinates, the answer is yes: you can choose any basis in the domain $\,V\,$ and the range $\,W\,$ and give a explicit formula for any map wrt these basis' coordinates in both spaces

Added Take the $\,3-$dimensional vector space $$\mathcal{P}_2:=\{f(x)\in\mathbb{R}[x]\;|\; \deg f\leq 2\}$$ First problem: what is "the standard basis" here? Let's take the basis $\,\{1,x,x^2\}\,$ and the derivation linear map $$D:\,\mathcal P_2\to\mathcal P_2\,\,,\,D(f(x)):=f'(x)$$The matrix of $\,D\,$ wrt the above basis is $$\begin{align*}D(1)&=0=0\cdot 1 +0\cdot x+0\cdot x^2\\D(x)&=1=1\cdot 1+0\cdot x+0\cdot x^2\\D(x^2)&=2x=0\cdot 1+2\cdot x+0\cdot x^2\end{align*}\Longrightarrow [D]=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}$$

If you choose any other basis in the domain/range you'll get something else, of course.

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Could you give me an example of how (2) works? –  Robert S. Barnes Jun 17 '12 at 6:22
    
@Robert, I added to my question –  DonAntonio Jun 17 '12 at 10:14

Elements of $\mathrm{Hom}(V,W)$ are just all $F$-linear maps $\varphi: V \to W$. The set of those again form a vectorspace over $F$ where scalarmultiplication and addition are given poitwise, i.e., for $f$ and $g$ such maps define $(f+g)(v) := f(v) + g(v)$ and for $\lambda \in F$ define $(\lambda\cdot f)(v):= \lambda\cdot(f(v))$. All of this is independent of any basis you could choose.

To your second question I would like to say two things. It is one of the first things one shows about linear maps of vectorspaces that whenever you choose any basis of $V$ you can define a linear map by sending your basis elements to anything you want in $W$. (This is also referred to as "Vectorspaces are free"). In particular it does not have to be the standard basis if $V = F^n$.

What is very important is to note that if $V$ is any $F$-vectorspace then there is no such thing as a standard basis of $V$ in general. In particular the second part of your second question does not really make sense for vectorspaces which are not equal to $F^n$ (of course they are isomorphic, but for this you have to choose a basis!).

I hope this helps.

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The other solutions are already doing a good job, but I think there is a little confusion here about thinking of bases of $V$ and $F^n$.

There is no "standard basis" of $V$, they are all on equal footing. I think standard basis usually refers to the basis of $F^n$ consisting of vectors with exactly one $1$ in a given position.

Every time you pick a basis of $V$, that gives rise to an isomorphism with $F^n$ in which a basis element for $V$ goes to one of those "standard basis" elements of $F^n$. (The isomorphism I'm thinking of sends each element of $V$ to the coefficients that represent it in that basis.) Try to sort these two bases out!

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