3
$\begingroup$

Suppose we have linear transformations $T : V \to W$ for finite dimensional vector spaces $V$ and $W$.
We can certainly think of $\mathcal{L}(V,W)$ as a vector space of all those transformations (provided addition and multiplication by scalar of transformations are defined and satisfy certain axioms).

While a vector space doesn't really need a basis and a dimension to exist, is there any meaning for a basis and dimension of this vector space $\mathcal{L}(V,W)$? Suppose $V$ is $n$-dimensional and $W$ is $m$-dimensional, what would be the dimension of $\mathcal{L}(V,W)$?

I was trying to make an analogy for the matrix representation of a linear transformation $T:V \to W$ as simply an isomorphism from $\mathcal{L}(V,W)$ to $F^{n\times m}$, the vector space of all $n \times m$ matrices over the field $F$.

Any clue about those points ?

$\endgroup$
1
  • 1
    $\begingroup$ A good reference for this is Section 2.4 from Linear Algebra by Friedberg et al. $\endgroup$
    – user70962
    Commented Apr 24, 2013 at 20:06

2 Answers 2

2
$\begingroup$

In order:

Yes, there is meaning for basis and dimension for $L(V,W)$, and they have meaning for every other vector space for that matter.

The dimension is $\dim(V)*\dim(W)$.

Yes, you can show that $L(V,W)\cong M_{\dim(V),\dim(W)}(\Bbb F)$ in such a way that evaluation of these transformations corresponds to matrix multiplication.

$\endgroup$
1
$\begingroup$
  1. You are Given Two Vector Spaces V and W.

    Now Any finite Dimensional Vector space Is Defined By its basis set B and Its field F.

2.Hence V=d(A,F) and W=d(B,F) where A is the Basis set Of V and B is that of W and F is the common field Of the vector Space.(d represents that it is defined by the following Sets)

  1. Hence We have Fixed A, B ,And F to define V(m)and W(n).

4. We know that Any Linear Transformation between Two fixed Basis of V and W respectively can be uniquely represented by A Matrix(mXn). Also, ANY MATRIX IN F(mXn) REPRESENTS AN UNIQUE LINEAR TRANSFORMATION FROM V->W AS THE BASES ARE FIXED.

  1. HENCE, the Required vector space is just the set of all F(mXn). and its dimension is m*n.

Beautiful isn't it!!!!!!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .