### determinant

January 16, 2024 math and physics play , , , , ,

I started reviewing a book draft that mentions the adjoint in passing, but I’ve forgotten what I knew about the adjoint (not counting self-adjoint operators, which is different.) I do recall that adjoint matrices were covered in high school linear algebra (now 30+ years ago!), but never really used after that.

It appears that the basic property of the adjoint $$A$$ of a matrix $$M$$, when it exists, is
M A = \Abs{M} I,

so it’s proportional to the inverse, where the numerical factor is the determinant of that matrix. Let’s try to compute this beastie for 1D, 2D, and 3D cases.

## Simplest case: $$1 \times 1$$ matrix.

For a one by one matrix, say
M =
\begin{bmatrix}
m_{11}
\end{bmatrix},

the determinant is just $$\Abs{M} = m_11$$, so our adjoint is the identity matrix
A =
\begin{bmatrix}
1
\end{bmatrix}.

Not too interesting. Let’s try the 2D case.

## Less trivial case: $$2 \times 2$$ matrix.

For the 2D case, let’s define our matrix as a pair of column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2
\end{bmatrix},

and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{bmatrix}.

We seek solutions to a pair of vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} &= \Abs{M} \Be_2.
\end{aligned}

We can immediately solve either of these, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 } a_{11} + \lr{ \Bm_2 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Be_1 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{11} + \lr{ \Bm_1 \wedge \Bm_2 } a_{21} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_1 } \\
\lr{ \Bm_1 \wedge \Bm_2 } a_{12} + \lr{ \Bm_2 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Be_2 \wedge \Bm_2 } \\
\lr{ \Bm_1 \wedge \Bm_1 } a_{12} + \lr{ \Bm_1 \wedge \Bm_2 } a_{22} &= \Abs{M} \lr{ \Bm_1 \wedge \Be_2}.
\end{aligned}

Any wedge with a repeated vector is zero.
Provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 = \Abs{M} \Be_{12}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
a_{11} &= \begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} \\
a_{21} &= \begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} \\
a_{12} &= \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
a_{22} &= \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{aligned}

or
A =
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} \\
& \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix}
\end{bmatrix},

or
A_{ij} =
\epsilon_{ir}
\begin{vmatrix}
\Be_j & \Bm_r
\end{vmatrix},

where $$\epsilon_{ir}$$ is the completely antisymmetric tensor, and the Einstein summation convention is in effect (summation implied over any repeated indexes.)

### Check:

We should verify that expanding these determinants explicitly reproduces the usual representation of the 2D adjoint:
\begin{aligned}
\begin{vmatrix} \Be_1 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 1 & m_{12} \\ 0 & m_{22} \end{vmatrix} = m_{22} \\
\begin{vmatrix} \Bm_1 & \Be_1 \end{vmatrix} &= \begin{vmatrix} m_{11} & 1 \\ m_{21} & 0 \end{vmatrix} = -m_{21} \\
\begin{vmatrix} \Be_2 & \Bm_2 \end{vmatrix} &= \begin{vmatrix} 0 & m_{12} \\ 1 & m_{22} \end{vmatrix} = -m_{12} \\
\begin{vmatrix} \Bm_1 & \Be_2 \end{vmatrix} &= \begin{vmatrix} m_{11} & 0 \\ m_{21} & 1 \end{vmatrix} = m_{11},
\end{aligned}

or
A =
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix}.

Multiplying everything out should give us determinant weighted identity
\begin{aligned}
M A
&=
\begin{bmatrix}
m_{11} & m_{12} \\
m_{21} & m_{22}
\end{bmatrix}
\begin{bmatrix}
m_{22} & -m_{12} \\
-m_{21} & m_{11}
\end{bmatrix} \\
&=
\lr{ m_{11} m_{22} – m_{12} m_{21} }
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
&= \Abs{M} I,
\end{aligned}

as expected.

## 3D case: $$3 \times 3$$ matrix.

For the 3D case, let’s also define our matrix as column vectors
M =
\begin{bmatrix}
\Bm_1 & \Bm_2 & \Bm_3
\end{bmatrix},

and let’s write the adjoint out in full in coordinates as
A =
\begin{bmatrix}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{bmatrix}.

This time, we seek solutions to three vector equations
\begin{aligned}
\Bm_1 a_{11} + \Bm_2 a_{21} + \Bm_3 a_{31} &= \Abs{M} \Be_1 \\
\Bm_1 a_{12} + \Bm_2 a_{22} + \Bm_3 a_{32} &= \Abs{M} \Be_2 \\
\Bm_1 a_{13} + \Bm_2 a_{23} + \Bm_3 a_{33} &= \Abs{M} \Be_3,
\end{aligned}

and can immediately solve, once again, by taking wedge products, yielding
\begin{aligned}
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{11} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Be_1 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{21} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Be_1 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{11} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{21} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{31} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_1 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{12} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Be_2 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{22} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Be_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{12} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{22} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{32} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_2 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{13} + \lr{ \Bm_2 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_3 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Be_3 \wedge \Bm_2 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_1 \wedge \Bm_3 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{23} + \lr{ \Bm_1 \wedge \Bm_3 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Be_3 \wedge \Bm_3 \\
\lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_1 }a_{13} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_2 }a_{23} + \lr{ \Bm_1 \wedge \Bm_2 \wedge \Bm_3 }a_{33} &= \Abs{M} \Bm_1 \wedge \Bm_2 \wedge \Be_3,
\end{aligned}

Any wedge with a repeated vector is zero.
Like before, provided the determinant is non-zero, we can divide both sides by $$\Bm_1 \wedge \Bm_2 \wedge \Bm_3 = \Abs{M} \Be_{123}$$ to find a single determinant for each element in the adjoint
\begin{aligned}
A &=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Be_1 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Be_3 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Bm_1 & \Bm_2 & \Be_1 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_2 \end{vmatrix} & \begin{vmatrix} \Bm_1 & \Bm_2 & \Be_3 \end{vmatrix}
\end{bmatrix} \\
&=
\begin{bmatrix}
\begin{vmatrix} \Be_1 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_2 & \Bm_3 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_2 & \Bm_3 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_3 & \Bm_1 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_3 & \Bm_1 \end{vmatrix} \\
& & \\
\begin{vmatrix} \Be_1 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_2 & \Bm_1 & \Bm_2 \end{vmatrix} & \begin{vmatrix} \Be_3 & \Bm_1 & \Bm_2 \end{vmatrix}
\end{bmatrix},
\end{aligned}

or
A_{ij} = \frac{\epsilon_{irs}}{2!} \begin{vmatrix} \Be_j & \Bm_r & \Bm_s \end{vmatrix}.

Observe that the inclusion of the $$\Be_j$$ column vector in this determinant, means that we really need only compute a $$2 \times 2$$ determinant for each adjoint matrix element. That is

A_{ij} = \frac{(-1)^j \epsilon_{irs}\epsilon_{jab}}{(2!)^2}
\begin{vmatrix}
m_{ar} & m_{as} \\
m_{br} & m_{bs}
\end{vmatrix}
.

This looks a lot like the usual minor/cofactor recipe, but written out explicitly for each element, using the antisymmetric tensor to encode the index alternation. It’s worth noting that there may be an error or subtle difference from the usual in my formulation, since wikipedia defines the adjoint as the transpose of the cofactor matrix, see: [1].

## General case: $$n \times n$$ matrix.

It appears that if we wanted an induction hypotheses for the general $$n > 1$$ case, the $$ij$$ element of the adjoint matrix is likely
\begin{aligned}
A_{ij} &= \frac{\epsilon_{i s_1 s_2 \cdots s_{n-1}}}{(n-1)!} \begin{vmatrix} \Be_j & \Bm_{s_1} & \Bm_{s_2} & \cdots & \Bm_{s_{n-1}} \end{vmatrix} \\
&= \frac{(-1)^j \epsilon_{i r_1 r_2 \cdots r_{n-1}} \epsilon_{j s_1 s_2 \cdots s_{n-1}} }{\lr{(n-1)!}^2}
\begin{vmatrix}
m_{r_1 s_1} & \cdots & m_{r_1 s_{n-1}} \\
\vdots & & \vdots \\
m_{r_{n-1} s_{1}} & \cdots & m_{r_{n-1} s_{n-1}}
\end{vmatrix}.
\end{aligned}

I’m not going to try to prove this, inductively or otherwise.

# References

[1] Wikipedia contributors. Minor (linear algebra) — Wikipedia, the free encyclopedia, 2023. URL https://en.wikipedia.org/w/index.php?title=Minor_(linear_algebra)&oldid=1182988311. [Online; accessed 16-January-2024].

## Static load with two forces in a plane, solved a few different ways.

There’s a class of simple statics problems that are pervasive in high school physics and first year engineering classes (for me that CIV102.)  These problems are illustrated in the figures below. Here we have a static load under gravity, and two supporting members (rigid beams or wire lines), which can be under compression, or tension, depending on the geometry.

The problem, given the geometry, is to find the magnitudes of the forces in the two members. The equation to solve is of the form
\label{eqn:twoForceStaticsProblem:20}
\BF_s + \BF_r + m \Bg = 0.

The usual way to solve such a problem is to resolve the forces into components. We will do that first here as a review, but then also solve the system using GA techniques, which are arguably simpler or more direct.

## Solving as a conventional vector equation.

If we were back in high school we could have written our forces out in vector form
\label{eqn:twoForceStaticsProblem:160}
\begin{aligned}
\BF_r &= f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \\
\BF_s &= f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } \\
\Bg &= g \Be_1.
\end{aligned}

Here the gravitational direction has been pointed along the x-axis.

Our equation to solve is now
\label{eqn:twoForceStaticsProblem:180}
f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } + f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } + m g \Be_1 = 0.

This we can solve as a set of scalar equations, one for each of the $$\Be_1$$ and $$\Be_2$$ directions
\label{eqn:twoForceStaticsProblem:200}
\begin{aligned}
f_r \cos\alpha + f_s \cos\beta + m g &= 0 \\
f_r \sin\alpha + f_s \sin\beta &= 0.
\end{aligned}

Our solution is
\label{eqn:twoForceStaticsProblem:220}
\begin{aligned}
\begin{bmatrix}
f_r \\
f_s
\end{bmatrix}
&=
{\begin{bmatrix}
\cos\alpha & \cos\beta \\
\sin\alpha & \sin\beta
\end{bmatrix}}^{-1}
\begin{bmatrix}
– m g \\
0
\end{bmatrix} \\
&=
\inv{
\cos\alpha \sin\beta – \cos\beta \sin\alpha
}
\begin{bmatrix}
\sin\beta & -\cos\beta \\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}
– m g \\
0
\end{bmatrix} \\
&=
\frac{ m g }{ \cos\alpha \sin\beta – \cos\beta \sin\alpha }
\begin{bmatrix}
-\sin\beta \\
\sin\alpha
\end{bmatrix} \\
&=
\frac{ m g }{ \sin\lr{ \beta – \alpha } }
\begin{bmatrix}
-\sin\beta \\
\sin\alpha
\end{bmatrix}.
\end{aligned}

We have to haul out some trig identities to make a final simplification, but find a solution to the system.

Another approach, is to take cross products with the unit force direction.  First note that
\label{eqn:twoForceStaticsProblem:240}
\begin{aligned}
\lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta }
&=
\Be_3 \lr{
\cos\alpha \sin\beta – \sin\alpha \cos\beta
} \\
&=
\Be_3 \sin\lr{ \beta – \alpha }.
\end{aligned}

If we take cross products with each of the unit vectors, we find
\label{eqn:twoForceStaticsProblem:260}
\begin{aligned}
f_r \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } + m g \Be_1 \cross \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } &= 0 \\
f_s \lr{ \Be_1 \cos\beta + \Be_2 \sin\beta } \cross \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } + m g \Be_1 \cross \lr{ \Be_1 \cos\alpha + \Be_2 \sin\alpha } &= 0,
\end{aligned}

or
\label{eqn:twoForceStaticsProblem:280}
\begin{aligned}
\Be_3 f_r \sin\lr{ \beta – \alpha } + m g \Be_3 \sin\beta &= 0 \\
-\Be_3 f_s \sin\lr{ \beta – \alpha } + m g \Be_3 \sin\alpha &= 0.
\end{aligned}

After cancelling the $$\Be_3$$’s, we find the same result as we did solving the scalar system. This was a fairly direct way to solve the system, but the intermediate cross products were a bit messy. We will try this cross product using the wedge product. Switching from the cross to the wedge, by itself, will not make things any simpler or more complicated, but we can use the complex exponential form of the unit vectors for the forces, and that will make things simpler.

## Geometric algebra setup and solution.

As usual for planar problems, let’s write $$i = \Be_1 \Be_2$$ for the plane pseudoscalar, which allows us to write the forces in polar form
\label{eqn:twoForceStaticsProblem:40}
\begin{aligned}
\BF_r &= f_r \Be_1 e^{i\alpha} \\
\BF_s &= f_s \Be_1 e^{i\beta} \\
\Bg &= g \Be_1,
\end{aligned}

Our equation to solve is now
\label{eqn:twoForceStaticsProblem:60}
f_r \Be_1 e^{i\alpha} + f_s \Be_1 e^{i\beta} + m g \Be_1 = 0.

The solution for either $$f_r$$ or $$f_s$$ is now trivial, as we only have to take wedge products with the force direction vectors to solve for the magnitudes.  That is
\label{eqn:twoForceStaticsProblem:80}
\begin{aligned}
f_r \lr{ \Be_1 e^{i\alpha} +} \wedge \lr{ \Be_1 e^{i\beta} } + m g \Be_1 \wedge \lr{ \Be_1 e^{i\beta} } &= 0 \\
f_s \lr{ \Be_1 e^{i\beta} +} \wedge \lr{ \Be_1 e^{i\alpha} } + m g \Be_1 \wedge \lr{ \Be_1 e^{i\alpha} } &= 0.
\end{aligned}

Writing the wedges as grade two selections, and noting that $$e^{i\theta} \Be_1 = \Be_1 e^{-i\theta }$$, we have
\label{eqn:twoForceStaticsProblem:100}
\begin{aligned}
f_r &= – m g \frac{ \gpgradetwo{\Be_1^2 e^{i\beta}} }{ \gpgradetwo{ \Be_1^2 e^{-i\alpha} e^{i\beta} } } = – m g \frac{ \sin\beta }{ \sin\lr{ \beta – \alpha } } \\
f_s &= – m g \frac{ \gpgradetwo{\Be_1^2 e^{i\alpha}} }{ \gpgradetwo{ \Be_1^2 e^{-i\beta} e^{i\alpha} } } = m g \frac{ \sin\alpha }{ \sin\lr{ \beta – \alpha } }.
\end{aligned}

The grade selection a unit pseudoscalar factor in both the denominator and numerator, which cancelled out to give the final scalar result.

## As a complex variable problem.

Observe that we could have reframed the problem as a multivector problem by left multiplying \ref{eqn:twoForceStaticsProblem:60} by $$\Be_1$$ to find
\label{eqn:twoForceStaticsProblem:120}
f_r e^{i\alpha} + f_s e^{i\beta} + m g = 0.

Alternatively, we could have written the equations this way directly as a complex variable problem.

We can now solve for $$f_r$$ or $$f_s$$ by multiplying by the conjugate of one of the complex exponentials. That is
\label{eqn:twoForceStaticsProblem:140}
\begin{aligned}
f_r + f_s e^{i\beta} e^{-i\alpha} + m g e^{-i\alpha} &= 0 \\
f_r e^{i\alpha} e^{-i\beta} + f_s + m g e^{-i\beta} &= 0.
\end{aligned}

Selecting the bivector part of these equations (if interpreted as a multivector equation), or selecting the imaginary (if interpreting as a complex variables equation), will eliminate one of the force magnitudes from each equation, after which we find the same result.

This last approach, treating the problem as either a complex number problem (selecting imaginaries), or multivector problem (selecting bivectors), seems the simplest. We have no messing cross products, nor do we have to haul out the trig identities (the sine difference in the denominator comes practically for free, as it did with the wedge product method.)

## Update to old phy356 (Quantum Mechanics I) notes.

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.