Undergrad Change of Basis and Unitary Transformations

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The discussion focuses on the relationship between two orthonormal basis sets in a vector space, denoted as {v_i} and {w_i}. It explores how the transformation between these bases can be expressed using a unitary matrix U, where the old basis vectors can be represented in terms of the new basis through a matrix S, with U being the inverse of S. Participants clarify that the equations involving basis vectors can be interpreted as matrix multiplications, requiring conventions for representing these vectors as column vectors. The conversation emphasizes the need to define U correctly, ensuring it reflects the transformation from the v-basis to the w-basis. Ultimately, the relationship between the two bases is encapsulated in the equation w = Uv, illustrating how the transformation operates.
devd
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Say, we have two orthonormal basis sets ##\{v_i\}## and ##\{w_i\}## for a vector space A.

Now, the first (old) basis, in terms of the second(new) basis, is given by, say,

$$v_i=\Sigma_jS_{ij}w_j,~~~~\text{for all i.}$$

How do I explicitly (in some basis) write the relation, ##Uv_i=w_i##, for some unitary matrix, ##U##?

What is the relation between the matrix formed by the numbers ##S_{ij}## and ##U##?
 
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The S_{ij} form a matrix. U is its inverse.
 
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mathman said:
The S_{ij} form a matrix. U is its inverse.

How do I see this explicitly?
 
devd said:
How do I see this explicitly?

Can you see how the simultaneous equations

$$v_i=\Sigma_jS_{ij}w_j,~~~~\text{for all i.}$$

can be expressed as an equation involving matrices?

It's the same idea as expressing

1) ##ax + by = c##
2) ##dx + ey = f##

as

##\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c \\f \end{pmatrix}##
 
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Stephen Tashi said:
Can you see how the simultaneous equations
can be expressed as an equation involving matrices?

It's the same idea as expressing

1) ##ax + by = c##
2) ##dx + ey = f##

as

##\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c \\f \end{pmatrix}##

Yes, but where I got confused was that the entries of the column vectors themselves are vectors.
$$v_j=\sum_jS_{ij}w_i,~~~~\text{for all}~ j.$$
The ##v_j~'s ## and ##w_i~'s## in the equation above are not numbers but vectors.

But, I think, I have figured it out.

Let, U be a linear transformation on the vector space A, and ##\{v_j\}## be a basis. Then,
$$Uv_j=\sum_iU_{ij}v_i$$
Now, we define
$$\sum_iU_{ij}v_i=w_j\\ \implies v_j=U^{-1}w_j$$

Now, if we write the old basis vectors ##\{v_j\}## in terms of the new basis,
$$v_j=\sum_iS_{ij}w_i=Sw_j$$
Comparing the two equations for ##v_j##, we see that ##S=U^{-1}.##

In fact, we can write the the operators, ##U## and ##S## as outer products of the basis vectors,
$$U=\sum_j v_j\otimes v_j\\ S=\sum_i w_i \otimes w_i.$$

Do you think this makes sense? Or are there still some lacunae in my understanding?
 
devd said:
Yes, but where I got confused was that the entries of the column vectors themselves are vectors.
$$v_j=\sum_jS_{ij}w_i,~~~~\text{for all}~ j.$$
You need to have a "##v_i##" instead of a "##v_j##" on the left hand side of that equation.

Yes, the terms involved in that sum , ##S_{i,j} w_i##, are vectors, but we aren't yet dealing with any representation of ##w_i## as a vector of numbers. So, effectively, ##w_i## plays the role of a variable, just like the typical uses of "##x##" and "##y##".

Let, U be a linear transformation on the vector space A, and ##\{v_j\}## be a basis. Then,
$$Uv_j=\sum_iU_{ij}v_i$$
Yes, those equations ( considering all ##i##) define a linear transformation ##U## expressed in the ##v##-basis.

To interpret each equation as involving a matrix multiplication ##U v_j = b_j##, we must adopt some convention about how ##v_j## is represented as a column vector. Using "##v_j##" as ambiguous notation, we could say that ##v_j## represents a column vector of "variables" that are all zeroes except for the variable "##v_j##" in its ##j##th entry. Or we could say that ##v_j## ( in ##v##-basis) will be a column vector with a 1 in the ##j##th entry of the column and zeroes elsewhere. So interpreting the left hand side of the equation as matrix multiplication does involve a some convention about how ##v_j## is represented as a column vector.

We must also adopt a similar convention about the column vector on the right hand side. We might say it represents a linear combination of vectors from the ##v##-basis and that the ##k##th entry of the column vector on the right side the equation represents the coefficient of ##v_k## in the linear combination.
Now, we define
$$\sum_iU_{ij}v_i=w_j$$

It isn't clear what you are defining. Are you defining ##U## or are you defining the ##w_j## ?

If we take the ##w##-basis as given, you are defining ##U## as a matrix whose ##j##th column consists of the coefficients needed to express ##w_j## as a linear combination of vectors in the ##v##-basis.

However, the usual way to express the ##w##-basis in terms of the ##v##-basis in matrix form would be to consider the column vector ##w = (w_1,w_2,w_3,..)## as the ##w##-basis expressed as a vector of "variables", and the column vector ##w = (v_1,v_2,v_3,...)## as the ##v##-basis expressed a vector of variables, and to define the matrix ##U## to satisfy ##w = Uv##.

This definition would imply ##w_i = \sum_j U_{i,j} v_j ##, so the summation is over the column index "##j##" instead of the row index "##i##".
 
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