## introduction

Let me pick up an old problem. The mathematics comes from the two posts “transpose matrix and adjoint operator” part 1 & part 2. The problem itself comes from the schur’s lemma post.

Upon further reflection, I am going to change the problem a little bit. Do not expect to see the same answers as before.

I am also going to work it twice, assuming that we are given different information as our starting point, but I’ll do it for the very same problem.

As I have said, the appropriate question for an introduction to ABO blood groups is: Can your mother donate blood to you? Until you can answer that question, you’re missing something about blood groups.

This example is not that good, but it does tie together the following concepts and computations:

• the transition matrix;
• the attitude matrix;
• the change of basis formula;
• the reciprocal basis;
• the transpose of a matrix;
• the adjoint of a linear operator;
• the matrix of the adjoint operator.

This example is intended as a guide to and a review of my relevant posts, and will not repeat very much of the explanations in them.

Well, I have found that the explanations are growing.

What is the problem? Given a matrix and a non-orthonormal basis, find the matrix of the adjoint operator with respect to the non-orthonormal basis.

What is the key? We only know one way to find the matrix of the adjoint operator: take the transpose of some matrix.

Here is the general case: given a matrix X with respect to a basis E, we can transpose it, getting $X^T\$. But that is the matrix of the adjoint with respect to the dual basis F. To get the matrix with respect to the basis E, we must do a change-of-basis to get from $X^T\$ with respect to to F to, say, $Y^T\$ with respect to E.

There is, however, a very special case: if E is an orthonormal basis, then it is its own reciprocal basis, F = E, and we were done as soon as we computed the transpose, becuse $X^T\$ is the matrix of the adjoint with respect to E.

The general problem, then, is: given a matrix, find the matrix of its adjoint with respect to a non-orthonormal basis. We will consider two possibiities: first, that the original basis is orthonormal; second, that it is not.

I would like to make one final point before we dive in. There is a fundamental truth here, regardless of the precise definition of the adjoint operator. The fact is that two matrices A and $A^T\$ represent two different but related linear operators (with respect to an orthonormal basis); but if we transform those matrices to a non-orthonormal basis, the two results will not be the transposes of each other.

In other words, the relationship between two linear operators which is represented by a matrix and its transpose, holds only in an orthonormal basis.

## first form of the question

First, let us suppose we are given a matrix A with respect to an orthonormal basis; and a description of a non-orthonormal basis.

Here is the given matrix A:

$A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 2\end{array}\right)$

(This much is the same as before; this is the same matrix as in the previous posts.)

Here is the given non-orthonormal basis $\{f_1, f_2\}\$ specified by an attitude matrix…

$att = \left(\begin{array}{cc} 1 & -1 \\ 1 & 0\end{array}\right)$

Find the matrix, which I will call $C^T\$, of the adjoint operator with respect to the non-orthonormal basis.

Easy enough, once you’re used to it: we have A with respect to an orthonormal basis, so we form its transpose $A^T\$. This is the matrix of the adjoint with respect to the orthonormal basis. Now we just need to transform $A^T\$ to the non-orthonormal basis.

We want the transition matrix P, which is the transpose of the attitude matrix: $P = {att}^T\$.

$P = \left(\begin{array}{cc} 1 & 1 \\ -1 & 0\end{array}\right)$

One form of the change-of-basis formula tells us how to find the matrix B with respect to the new basis corresponding to A with respect to the original:

$B = P^{-1} A P\$,

so we get

$B = \left(\begin{array}{cc} 2 & 0 \\ -2 & 1\end{array}\right)$

That is, A and B are matrix representations, with respect to two bases, of a single linear operator. We don’t actually need the matrix B, but it serves for displaying the general change-of-basis formula for from A to B. And we will want to contrast it with $C^T\$.

We have been asked to find the matrix $C^T\$ corresponding to $A^T\$ with respect to a new basis. But that uses the change-of-basis formula, with $A^T\$ and $C^T\$ in place of A and B respectively:

$C^T = P^{-1} A^T P\$,

so we get

$C^T = \left(\begin{array}{cc} 1 & -1 \\ 0 & 2\end{array}\right)$

What we see (as we should expect) is that B and $C^T\$ are not the transposes of each other.

We have, then, 4 matrices. A and $A^T\$ represent a linear operator and its adjoint with respect to an orthornormal basis. B and $C^T\$ represent the same linear operator and its adjoint with respect to the non-orthonormal basis $\{f_1, f_2\}\$.

(In principle, of course, we only needed three of those matrices: A, $A^T\$, and $C^T\$; but it may be informative to compute B.)

## second form of the question first of three solutions

We are given the matrix B, representing a linear operator with respect to a non-orthonormal basis…

$B = \left(\begin{array}{cc} 2 & 0 \\ -2 & 1\end{array}\right)$

and we were given the attitude matrix describing that non-orthonormal basis…

$att = \left(\begin{array}{cc} 1 & -1 \\ 1 & 0\end{array}\right)$

We want to find the matrix $C^T\$ of the adjoint operator with respect to the non-orthonormal basis.

The quickest way to do it is to transform B to the orthonormal basis, which gives us A. We still use the transition matrix $P = {att}^T \$

$P = \left(\begin{array}{cc} 1 & 1 \\ -1 & 0\end{array}\right)$

and the change of basis formula (but note that we’re going in the other direction, to the orthonormal basis instead of from it)…

$A = P B P^{-1}\$.

so we get

$A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 2\end{array}\right)$

We have just reduced this question to the previous problem: now we can get $C^T\$ by transforming $A^T\$ to the non-orthonormal basis…

$C^T = P^{-1} A^T P\$.

We’re done. We could stop here. Maybe we should. But it’s nice to know more than one way to do something, because then we can check our work.

## second of three solutions

Alternatively, we could use the reciprocal basis.

This gets a little messy only because we’re introducing a 3rd basis, so we get 2 more matrices.

We transpose B, and we get the matrix of the adjoint operator with respect to the reciprocal basis:

$B^T = \left(\begin{array}{cc} 2 & -2 \\ 0 & 1\end{array}\right)\$.

But we’ve only just begun. We don’t want the adjoint with respect to the reciprocal basis; we want it with respect to the given non-orthonormal basis. It’s just that, ultimately, we have to take the transpose with respect to some basis; it’s the only way to compute the matrix of the adjoint. Then the question

how do we transform from the reciprocal basis (where we have $B^T\$) to the non-orthonormal basis where we want $C^T\$?

becomes

what is the transition matrix from the reciprocal basis to the basis?

And that leads to the question: exactly what is our reciprocal basis?

I could simply say (!) that the attitude matrix for the reciprocal basis is the inverse of the transition matrix P for the non-orthonormal basis; that is, the transition matrix Q for the reciprocal basis is the inverse transpose of P:

$Q = P^{-T}\$,

$Q = \left(\begin{array}{cc} 0 & 1 \\ -1 & 1\end{array}\right)$

But I think it’s only fair to admit that I don’t remember the description of the attitude matrix for the reciprocal basis. What I have in my head is a matrix equation, which says that the product of the attitude matrix Q for the reciprocal basis with the transition matrix P for the given basis, is the identity:

$Q^T P = I\$.

(And that matrix equation simply says that the dot products of the reciprocal basis vectors – the rows of $Q^T\$ – with the given non-orthonormal basis vectors – the columns of P – are either 0 or 1.)

Then I solve for Q:

$Q = P^{-T}\$.

Of course Q is what I said it was, but the point is that I remember the derivation rather than the answer. Fortunately it’s a very short and easy derivation, so I see the answer almost immediately….

Having Q, we can transform $B^T\$ to (not from) the orthonormal basis…

$D^T = Q B^T Q^{-1}\$,

so we get

$D^T = \left(\begin{array}{cc} 1 & 0 \\ 1 & 2\end{array}\right)$

and then we could transform $D^T\$ to the non-orthonormal basis:

$C^T = P^{-1} D^T P\$,

$C^T = \left(\begin{array}{cc} 1 & -1 \\ 0 & 2\end{array}\right)$

That is, as it should be, the same answer.

## third of three solutions

And the second solution shows us another way to have done it. From

$C^T = P^{-1} D^T P\$

and

$D^T = Q B^T Q^{-1}\$

we get

$C^T = P^{-1} Q B^T Q^{-1} P\$,

which we write as

$C^T = R^{-1} B^T R\$,

with R defined as

$R = Q^{-1} P\$.

By writing out the algebra first, we can dispense with explicit the computation and transformation of $D^T\$.

In other words, the transition matrix R from the reciprocal basis to the non-orthonormal basis is given by $R = Q^{-1} P\$. Equivalently, the transition matrix from the non-orthonormal basis to its reciprocal basis is given by $P^{-1} Q\$.

(But even if we use R, we are implicitly going through the original orthonormal basis.)