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I'm trying to learn suffix notation (both to prove results in linear algebra and for application in vector calculus).

As an exercise, I wanted to use it to prove that the determinant of a product of two matrices is equal to the product of their determinants, i.e. for $\underline{A},\, \underline{B},\, \underline{C}$, $3\times 3$ matrices where $C_{pq} = A_{pr}B_{rq}$, I'd like to show that:

$$\det(\underline{A}\,\underline{B}) = \det(\underline{A}) \det(\underline{B})$$


$\color{blue}{\textbf{Here is what I have tried so far}}$:

$$\det{(\underline{C})} = \det(\underline{A}\,\underline{B}),$$

Now, the determinant in suffix notation would be:

$$\begin{align*} \epsilon_{ijk} C_{1i} C_{2j} C_{3k} & = \epsilon_{ijk} \, A_{1p}B_{pi} \, A_{2q}B_{qj} \, A_{3r}B_{rk}\\ & = \epsilon_{ijk} \, A_{1p}A_{2q}A_{3r} \, B_{pi}B_{qj}B_{rk} \end{align*}$$

$\color{blue}{\textbf{At this point, I couldn't find any way forward}}$. I tried reading up on tensors, to build my intuition, but after hours spent with several different texts, I was only more confused than before, so I tried to focus on the formalism, itself.


$\color{green}{\textbf{Eventually, I thought to try it from the other side}}$, i.e. starting with $\det(\underline{A}) \det(\underline{B})$, and (after spending more time than I care to admit) I managed this:

$$ \begin{align*} \det(\underline{A}) \det(\underline{B}) %% = & \epsilon_{ijk} \, A_{1i}A_{2j}A_{3k} \, \epsilon_{pqr}B_{1p}B_{2q}B_{3r} \\[5pt] %% = & \epsilon_{ijk} \epsilon_{pqr} \, A_{1i}A_{2j}A_{3k} \, B_{1p}B_{2q}B_{3r} \\[5pt] %% = & \bigl( \delta_{ip}(\delta_{jq}\delta_{kr} - \delta_{jr}\delta_{kq}) % + \delta_{iq}(\delta_{jr}\delta_{kp} - \delta_{jp}\delta_{kr}) % + \delta_{ir}(\delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}) \bigr) A_{1i}A_{2j}A_{3k} \, B_{1p}B_{2q}B_{3r} \\[5pt] %% = & (\delta_{jq}\delta_{kr} - \delta_{jr}\delta_{kq}) A_{1i}A_{2j}A_{3k} \, B_{1{\color{red}i}}B_{2q}B_{3r}\\ % & \quad + (\delta_{jr}\delta_{kp} - \delta_{jp}\delta_{kr}) A_{1i}A_{2j}A_{3k} \, B_{1p}B_{2{\color{red}i}}B_{3r}\\ % & \qquad + (\delta_{jp}\delta_{kq} - \delta_{jq}\delta_{kp}) A_{1i}A_{2j}A_{3k} \, B_{1p}B_{2q}B_{3{\color{red}i}}\\[5pt] %% = & \quad A_{1i}A_{2j}A_{3k} B_{1{\color{red}i}} (B_{2j}B_{3k} - B_{2k}B_{3j}) \\ & + A_{1i}A_{2j}A_{3k} B_{2{\color{red}i}} (B_{1k}B_{3j} - B_{1j}B_{3k}) \\ & + A_{1i}A_{2j}A_{3k} B_{3{\color{red}i}} (B_{1j}B_{2k} - B_{1k}B_{2j}) \end{align*} $$

$\color{green}{\textbf{I } think \textbf{ this last expression can be written as}}$:

$$\begin{align*} \det(\underline{A} \, \underline{B}) %% & = A_{1i}A_{2j}A_{3k} \, \epsilon_{pqr}B_{pi}B_{qj}B_{rk} \\[5pt] %% & = \epsilon_{pqr} \, A_{1i}B_{pi}\, A_{2j}B_{qj} \, A_{3k}B_{rk}; \end{align*}$$

however, it's possible my reasoning is incorrect (I'm still very uncertain about the formalism...).


$\color{red}{If \textbf{ the above is correct}}$, it seems to look like $\det(\underline{A} \underline{B}^T)$, which suggests that if I had started off by noting that:

$$ \det(\underline{A})\, \det(\underline{B}) = \det(\underline{A}) \, \det(\underline{B}^T)$$

$\color{red}{\textbf{I would end up with the correct final expression}}$. However, (if all my reasoning to date is correct) my question(s) are:

  1. is it possible to see that $$\epsilon_{pqr} \, A_{1i}B_{{\color{blue}p}i}\, A_{2j}B_{{\color{blue}q}j} \, A_{3k}B_{{\color{blue}r}k} = \epsilon_{pqr} \, A_{1i}B_{i{\color{blue}p}}\, A_{2j}B_{j{\color{blue}q}} \, A_{3k}B_{k{\color{blue}r}}$$ directly (i.e. without having to think about the interpretation as the product of determinants etc; simply via identities or application of the definitions)? If so, how?

  2. is there some way of doing this (using suffix notation) without having to expand the epsilon product $\epsilon_{ijk}\, \epsilon_{pqr}$, explicitly (all the resources I have read so far have only offered the identity in terms of the Kronecker deltas, nothing that relates the product to another epsilon)?

This is painfully long to write out, so would equally appreciate mere direction towards a good "workbook" with lots of problems (& answers or solutions) related to suffix notation. My hope is that if I first become comfortable with the formalism, I might later have a more success trying to understand the math.

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    $\begingroup$ What is the formula you are using for the determinant? I cannot make of your first line of working with $\epsilon_{ijk} C_{1i} C_{2j} C_{3k}$. If $C$ is an $n \times n$ matrix, the determinant should be a polynomial of degree $n$ in the $C_{ij}$, not of degree 3 as is written there. I would believe it for 3 by 3 matrices though. $\endgroup$ – Joppy Mar 8 '18 at 6:03
  • $\begingroup$ @Joppy indeed, your entirely correct; this just goes to show how green I still am with the notation (easily forgetting the dimensions I'm working with). That would mean one would need $\epsilon_{i_1 i_2 \dots i_n}$, correct? I will correct the question to say $3 \times 3$, because I am sufficiently confused by this relatively simple case, and am hoping that a firm understanding will help to later generalize to $n \times n$. Thank you for taking the time to read - this took forever to write out & has been on my mind for over a week! $\endgroup$ – Rax Adaam Mar 8 '18 at 16:00
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Part of the trouble is that you are using a weaker version of the determinant identity. You have $$\det(B) = \epsilon_{ijk}B_{1i}B_{2j}B_{3k},$$ which is of course true. But this singles out the rows by giving them concrete values, and it's best to avoid things like that. It's almost always better to work with abstract index notation than to work with the concrete components of the tensor, as you will see. Let us write the above identity as $$\det(B)\epsilon_{\ell m n}=\epsilon_{ijk}B_{\ell i}B_{mj}B_{nk},\tag{*}$$ which you can see is true simply because the determinant is alternating on rows and columns. Your version is simply the component where $\ell = 1$, $m = 2$ and $n = 3$. Now contract with the appropriate components of $A$ to get $$\begin{align}\det(A)\det(B)\epsilon_{rst} &= \det(B)\epsilon_{\ell m n}A_{r\ell}A_{sm}A_{tn}\\ &= \epsilon_{ijk}(A_{r\ell}B_{\ell i})(A_{sm}B_{mj})(A_{tn}B_{nk})\\ &= \det(AB)\epsilon_{rst}.\end{align}$$ A deeper reason for what's going on is that the Levi-Civita symbol is really representing the top level form of the exterior algebra, and identity $(*)$ is really the components for $$\det(B)\,e_1\wedge e_2 \wedge \cdots \wedge e_n = B(e_1) \wedge B(e_2) \wedge \cdots \wedge B(e_n).$$ By supressing the epsilon symbol in $(*)$, you're working with a single component of the top level form, which of course makes things very difficult.

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  • $\begingroup$ Thank you very much for this explanation. I can't pretend to follow anything of the "deeper reason" (yet!), but think I am basically comfortable with the formal approach (& hopefully will be able to use this idea elsewhere). Really appreciate your taking the time to read & reply! $\endgroup$ – Rax Adaam Mar 8 '18 at 18:41
  • $\begingroup$ was my reasoning correct, nonetheless? (It's okay if you didn't read through it - I know it's tedious! I just hope someone else will find it helpful!). $\endgroup$ – Rax Adaam Mar 8 '18 at 18:45

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