# I Change of Basis and Unitary Transformations

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1. Feb 17, 2017

### devd

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 $U$?

2. Feb 17, 2017

### mathman

The $$S_{ij}$$ form a matrix. U is its inverse.

3. Feb 17, 2017

### devd

How do I see this explicitly?

4. Feb 17, 2017

### Stephen Tashi

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

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

5. Feb 17, 2017

### devd

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?

6. Feb 18, 2017

### Stephen Tashi

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 "$y$".

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.

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 "$i$".