Category: Linear Algebra

Change of basis of a linear map

1. Coordinate maps

consider a random vector a. If the values of every entry are explicitly given, one could naturally identify it as a vector under the orthonormal basis. However, it could also been understood as the coordinate vector, which represents a vector under another basis by a coordinate map:

2. Change of basis of a linear map

Now we consider a linear map (represented by a matrix) A. It could be interpreted  as:

What if we want to change the basis of the objective vector space of f ? The following relation is obvious, if we want to find a matrix A’ representing the map after change of basis:

Where P is defined as a matrix that perform this basis change.

[Literature: Andre Lukas, Lecture note on Vectors and Matrices, University of Oxford]

3. The invariant map

Suppose we have a map that is invariant under any basis change, that is:

In other words, we would like to find an operator that commutes with any other operator on the same vector space V. Suppose now we have a vector x in V. We would like to find a non-trivial linear functional on x, so that we can define a linear map:

This is possible, for a functional in a n dimensional space can be express in to an (1 x n) matrix, so the left hand side can be expressed as (nx1)(1xn)(nx1) corresponding to v, f, and x respectively, and the former 2 combined and form a matrix. ( My reference mentioned ” Axiom of choice” with respect to finding non-trivial functional, and yet I have not understood it perfectly)

Then, the commutation relation implies:

Note that f(Tx), according to our pervious argument, should exist and independent on v. Thus, what we are doing consequently is that we have constructed linear maps P according to our need (that is, the vector v). In other words, we can assign every vector v in V a matrix P and they have to satisfy (6).

Then this inplies:

i.e. T is the scalar multiple of  the identity.

[Literature: Robert Isreal, https://math.stackexchange.com/questions/27808/a-linear-operator-commuting-with-all-such-operators-is-a-scalar-multiple-of-the]

 

Linear Functionals

1.Vectors

It is important to formalised the definition of a vector and a vector space. Here I shall pay no more attention on the this issue. But they are important as they characterised the condition that, for example, the scalar field accompanying with V, must satisfy.

1.Functionals

One should be very familiar with the definition of “linear” and “linear maps”(And from the definition we see linear maps can from a linear space: space of linear functions, as well). Very interestingly, we can show the following two important relations: for f: V→W:(See reference)

and that if ker= {0}, f: V→V is isomorphism.

We can get the following observation: consider a linear map with dim(imf) = 1 (Which, since vector spaces are defined base on a scalar field, this 1 dimensional space is the field itself. We name this special type of map linear functionals. Since we know that linear functions themselves form a vector space, we call it dual space V *(n,K) , this apply to linear functionals as well. Form (1) we know that one particular functional actually only act on one specific dimension, we conclude that there are only dim(V) types of linearly independent functionals. i.e. dim(V)=dim(V*). It follows that for basis {e} in V and basis {g} in V*, we naturally pair them together:

So that V and V* are putted in an equal status (symmetry).

Now, suppose we have f: V→W and g: W→F where g is a functional, it is easy to find the a functional in V*, namely, g( f ).  Then, we could also define another space of functionals h: W*→V* as h: g→f = h(g). 

We can also define a map i: V→V* and we can show the condition for it to be isomorphism. Since dimV = dim V* if we also know Ker(i) = 0 this should do it. Remember that for defining a dual space, it is necessary to recall that it represents a space of linear maps.  For any non-zero v, if w = f(v) is non zero (Note that the image of zero is always zero) the corresponding dual vector is (by our definition) non zero. (In other word, this map is not degenerate) However, we shall note that the proportionality between e and image of e by g is not restricted. We could try to more specify it:

Consider the effect on w, we also use it to define the inner product <,>: V x V→F:

[Reference: Geometry, Topology and Physics. M Nakahara(2003)]

Wronskian

1. Derivative of a determinant

Consider the determinant W(t) of a n×n matrix Y which each element is a function of t. Assume elements to be independent variables. Then we could write:

Where Cij are the corresponding cofactors. Thus we have:

Let’s define γi as the new matrices form by substituting the ith row with its derivative. Then we could write (2) in a tidier form:

2. Abel-Jacobi-Liouville identity

As we know, any system of linear ordinary equations can be extracted in to a single linear equation, namely:

And it is followed by that,

So we obseverve that a particular row of the derivative is the linear combination of the original rows, since for the kth row of Y, different elements on jth column are multiplied by the same factor Aik. So of course, each term on the right hand side of (3) will be W times the corresponding element of Aii.
Thus,

This has resulted in some interesting conclusions. For example if the solutions are independent for any point within the domin, they must be independent entirely.

[Literature: Pontryagain 1962, Chapter 3]