The vector space of linear transformations

[Bipolarity]

Consider the operation of multiplying a vector in ℝ^{n} by an m \times n matrix A. This can be viewed as a linear transformation from ℝ^{n} to ℝ^{m}. Since matrices under matrix addition and multiplication by a scalar form a vector space, we can define a "vector space of linear transformations" from ℝ^{n} to ℝ^{m}.

My question is whether this connection between linear transformations and transformation matrices exists for linear mappings to and from arbitrary vector spaces. So given some general n-dimensional vector space U and m-dimensional vector space W, can every linear mapping from U to W be viewed as multiplication by a m \times n transformation matrix ?

Or is there a linear transformation which cannot be viewed as multiplication by a transformation matrix?

BiP

 

[CompuChip]

Yes, because you can always reduce your general situation down to $\mathbb{R}^n$.

Given a general n-dimensional vector space U, you can choose a basis $\{ \hat e_1, \hat e_2, \ldots, \hat e_n \}$ (and without loss of generality you can let it be orthogonal, for convenience). Now consider the map $$\phi: U \to \mathbb{R}^n, u_1 \hat e_1 + \cdots + u_n \hat e_n \mapsto ( u_1, \ldots, u_n)$$.

 

[Bipolarity]

What is \{ u_{1},u_{2}...,u_{n} \} ?
And once you've reduced to ℝ^{n} and ℝ^{m}, is it necessarily the case that the linear transformation can be viewed as multiplication by a transformation matrix? Would we need to establish an isomorphism between them?BiP

 

[WannabeNewton]

Every linear map between two finite dimensional vector spaces can be represented as a matrix between the associated euclidean spaces. This is called the matrix representation of the linear map and is a standard topic in most linear algebra textbooks. See, for example, chapter 5 of Lang "Linear Algebra".

 

[CompuChip]

What is \{ u_{1},u_{2}...,u_{n} \} ?

They are the components of the vector in U. In the definition of the function I wrote an arbitrary element of U as $u_1 \hat e_1 + u_2 \hat e_2 + \cdots + u_n \hat e_n$ which I can do because $\{ \hat e_i \mid i = 1, \ldots, n \}$ is a basis.

Every linear map between two finite dimensional vector spaces can be represented as a matrix between the associated euclidean spaces.

To give you an idea: choose a basis $\{ \hat e_1, \hat e_2, \ldots, \hat e_n \}$ on $\mathbb{R}^n$ and $\{ \hat f_1, \hat f_2, \ldots, \hat f_m \}$ on $\mathbb{R}^m$. Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation. Then you can write
$$T \hat e_i = a_{i,1} \hat f_1 + \cdots + a_{i,m} \hat f_m = \sum_{j = 1}^m a_{ij} \hat f_j$$
The coefficients $a_{ij}$ form entries of an $n \times m$ matrix $A$. It is straightforward to check that multiplying $A$ with a vector which contains a 1 in position i and 0 in all others (e.g. (0, 0, ..., 0, 1, 0, 0, ...)) gives you the correct coefficients, so that A is the matrix representation of T.
Note that I make a distinction between the linear map T itself and the matrix A. Of course, you are free to choose another basis on $\mathbb{R}^n$ and/or $\mathbb{R}^m$, which will change the numbers in the matrix. The transformation T doesn't change though - this can be confusing at first! :)

Now if $S: U \to V$ is another transformation you can let $\phi: U \to \mathbb{R}^n$ be as I described earlier, mapping an element of U into a vector in $\mathbb{R}^n$ by picking out the components relative to some arbitrary basis. Let $\psi: V \to \mathbb{R}^m$ be a similar map for V. Now you can show that $\phi$ and $\psi$ are bijections and therefore for any u in U, you can write S(u) as $\psi^{-1} A \phi u$ for some suitable matrix A, which just represents a linear transformation $\mathbb{R}^n \to \mathbb{R}^m$.

If you don't follow the above paragraph, just think of it this way: if U is an n-dimensional vector space, you can see it as just $\mathbb{R}^n$ looking slightly different. So any map between U and an m-dimensional space V is secretly just a map from $\mathbb{R}^n$ to #\mathbb{R}^m##.

(PS: Note that things do get trickier if the spaces are no longer finite-dimensional.)