College Math Teaching

October 7, 2016

Now what is a linear transformation anyway?

Filed under: linear albegra, pedagogy — Tags: , — collegemathteaching @ 9:43 pm

Yes, I know, a linear transformation L: V \rightarrow W is a function between vector spaces such that L(V \oplus W) = L(V) \oplus L(W) and L(a \odot V) = a \odot L(V) where the vector space operations of vector addition and scalar multiplication occur in their respective spaces.

Previously, I talked about this classical example:

Consider the set R^+ = \{x| x > 0 \} endowed with the “vector addition” x \oplus y = xy where xy represents ordinary real number multiplication and “scalar multiplication r \odot x = x^r where r \in R and x^r is ordinary exponentiation. It is clear that \{R^+, R | \oplus, \odot \} is a vector space with 1 being the vector “additive” identity and 0 playing the role of the scalar zero and 1 playing the multiplicative identity. Verifying the various vector space axioms is a fun, if trivial exercise.

Then L(x) = ln(x) is a vector space isomophism between R^+ and R (the usual addition and scalar multiplication) and of course, L^{-1}(x) = exp(x) .

Can we expand this concept any further?

Question: (I have no idea if this has been answered or not): given any, say, non-compact, connected subset of R, is it possible to come up with vector space operations (vector addition, scalar multiplication) so as to make a given, say, real valued, continuous one to one function into a linear transformation?

The answer in some cases is “yes.”

Consider L(x): R^+ \rightarrow R^+ by L(x) = x^r , r any real number.

Exercise 1: L is a linear transformation.

Exercise 2: If we have ANY linear transformation L: R^+ \rightarrow R^+ , let L(e) = e^a .
Then L(x) = L(e^{ln(x)}) = L(e)^{ln(x)} = (e^a)^{ln(x)} = x^a .

Exercise 3: we know that all linear transformations L: R \rightarrow R are of the form L(x) = ax . These can be factored through:

x \rightarrow e^x \rightarrow (e^x)^a = e^{ax} \rightarrow ln(e^{ax}) = ax .

So this isn’t exactly anything profound, but it is fun! And perhaps it might be a way to introduce commutative diagrams.

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