# Change of basis

1. Feb 18, 2012

### matqkks

Why should we want to write a vector in Rn in other than standard basis?
A normal application of linear transformations in most textbooks is converting a given vector in standard basis to another basis. This is sometimes a tedious task. Why carry out this task?

2. Feb 22, 2012

### Jolb

There are many scenarios you'd want to change bases.

Here's one example: let's say you have an invertible real matrix M, and suppose that it is symmetric. We know from linear algebra that there is an orthogonal matrix D such that D-1MD is a diagonal matrix, where the diagonal entries are the eigenvalues of M. (D-1 is the inverse of D.) D-1MD is the representation of the matrix in the eigenvector basis, and its action on eigenvectors is to simply multiply them by the scalar eigenvalue.

That's a very abstract explanation, so I'll explain it in a more commonplace way.

Suppose we have M, a real, symmetric, invertible matrix (linear transformation). Let's call its dimension N, so it's an NxN matrix.

We know from linear algebra that there will be N linearly independent eigenvectors. For an eigenvector x there is a scalar v, called the eigenvalue, such that Mx = vx. Since there are N linearly independent eigenvectors, they form a basis for Rn.

Let's say we have a vector r with the standard basis representation:
$$[\vec{r}]_{standard-basis}=\begin{bmatrix} r_1\\ r_2\\ ...\\ r_N \end{bmatrix}$$
Because the eigenvectors $$\left \{ \vec{x_1}, \vec{x_2}, ... ,\vec{x_N} \right \}$$ form a basis, then for some coefficients c1, c2, ... cN,
$$\vec{r}=c_1\vec{x_1}+ c_2\vec{x_2}+ ... +c_N\vec{x_N}$$

This means that
$$[\vec{r}]_{eigenvector-basis}=\begin{bmatrix} c_1\\ c_2\\ ...\\ c_N \end{bmatrix}$$

Now we can easily compute the action of M on the vector r.

$$M\vec{r}=M(c_1\vec{x_1}+ c_2\vec{x_2}+ ... +c_N\vec{x_N})=c_1M\vec{x_1}+ c_2M\vec{x_2}+ ... +c_NM\vec{x_N} = c_1v_1\vec{x_1}+ c_2v_2\vec{x_2}+ ... +c_Nv_N\vec{x_N}$$

Notice we avoided doing any actual matrix multiplication. The only thing we had to do was express the vector r in the basis of eigenvectors.

It makes it quick and easy to compute the action of a linear transformation on a vector if the vector is expressed in the eigenbasis. This is very common in Quantum Mechanics.

Edit: Oops, you got me deveno. I forgot that M must by symmetric. Corrected the mistake.

Last edited: Feb 22, 2012
3. Feb 22, 2012

### Deveno

another example, is given by the space of all real polynomials, defined on the interval [-1,1]. the "standard" basis is {1,x,...,xn...}, where each polynomial is represented by the vector of its coefficients.

the thing is, there are different possible inner products definable on any given vector space. with the standard "dot" product, the above basis is orthogonal. if, however, the inner product is:

$$\int_{-1}^1 p(x)q(x) dx$$

then the above basis is no longer orthogonal, and one might prefer the following basis:

{1,x,(1/2)(3x2-1), (1/2)(5x3-3x), (1/8)(35x4-30x2+3),...Pn(x),...}

where Pn is the n-th legendre polynomial.

if one thinks of vectors as standing for "actual things" (with the necessary properties), it may well be that the different ways of "representing" these objects have different advantages/disadvantages. that is, reducing vectors to arrays of field elements (numbers) might depend on how we do this (the units involved, for example). the actual vectors involved in analyzing forces on an object don't actually change, just because we shift from "meters" to "feet", but the numbers DO. this comes into play in multi-variate calculus, where one might have to deal with how a change of coordinates affects an integral, even though the region being integrated doesn't change.

in many different branches of math, the "concrete" realization of an "abstract" structure, often cannot be done without making some "choice". in linear algebra, this choice is "choosing a basis". the numerical results obtained depend on the basis chosen. as Jolb pointed out, if a matrix is diagonalizable, choosing a basis wisely, greatly reduces the amount of arithmetic you have to do.

(i do want to point out one small error in the post above:

an invertible matrix is not necessarliy diagonalizable. for example:

[1 1]
[0 1] is invertible, but the eigenspace corresponding to the single eigenvalue 1 is deficient, because its dimension is only 1, so there is no eigenbasis).

4. Feb 22, 2012

### HallsofIvy

I would like to point out that, in an abstract vector space, there is NO "standard basis". You can, of course, represent any vector space in terms of Rn but that requires choosing a specific basis for the vectors space first.

5. Feb 22, 2012

### dx

A real vector V space does not come equipped with a natural basis. Once a basis is chosen, it induces a map from V into the space Rn of n-tuples of real numbers, i.e. a vector v is sent to (v1, ..., vn), where vi are the componets of v in this basis.

6. Feb 22, 2012

### Deveno

fair enough, but there are reasons for choosing the standard basis as standard in Fn, for any field F, because then (if we call this basis B):

[v]B = v

(the coordinates of the vector v in that basis form a coordinate vector which is v itself).

in much the same vein, choosing the standard basis for Matmxn(F), allows us to use coordinates that are the individual matrix entries.

in other words, for certain vector spaces, the standard basis is "invisible". this makes the "translation" part (mapping our "standard" basis to an "arbitrary" basis) carried by a linear algebra isomorphism: when we specify a basis for the vector space of real polynomials, we implicitly define this "change of basis" transformation, because we have no way of naming individual polynomials without assigning some basis (that is, saying which coefficients they have). there are some good reasons for putting all the difficulty in the "hom-set" (the linear transformations), the rank-nullity theorem is one of them.

put another way: we can prove the differentiation operator is a linear operator, without exhibiting a basis for the space of real polynomials, but we cannot actually compute a derivative without choosing a basis.

one sees this all the time in field theory: one can define F(a) abstractly as an extension of F, and as an extension it automatically becomes a vector space over F. but if you want to know: is this certain element of E (where E is some other extension of F) in F(a), you need a way to express elements of F(a) in terms of elements of F, and a.

or: in taking determinants...although the determinant is an invariant of a linear transformation, actually calculating one requires you choose a basis.

my apologies for being somewhat long-winded. my point is: there is very good reason for picking Fn as the canonical example of an n-dimensional vector space: it has a canonical basis. and vector spaces are "free objects": dimension completely determines a vector space (up to isomorphism).

i do agree, however, that when talking about vector spaces (as in teaching a class), one should underscore that a basis is indeed a choice, and we are free to make it arbitrarily (any LI spanning set will do). for many purposes, we don't need to know which one we picked.