2

Firstly, I will note that I have looked at many sources on Google, and also on Stack exchange, but have not been able to find a hint of how to go about proving this the way I am doing- or at least not one which I could connect with the symbols/notation I am using here. I think that the first answer in this post is quite similar, but I am not sure. I am not familiar with manipulating the symbols/notation that was used there.

So far I have:

$$(AB)_{ij}=p_{ij}=\sum _k (A)_{ik}(B)_{kj}$$

Then

$$\det(AB)=\sum_{j_1j_2...j_n}\epsilon _{j_1j_2...j_n}p_{1j_1}p_{2j_2}...p_{nj_n}$$

Where $\epsilon _{j_1j_2...j_n}$ is the Levi-Civita or epsilon tensor

$$\det(AB)=\sum_{j_1j_2...j_n}\epsilon _{j_1j_2...j_n}\sum_{k_1} (A)_{1k_1}(B)_{k_1j_1}...\sum_{k_n} (A)_{nk_n}(B)_{k_nj_n}$$

$$=\sum_{j_i,k_i}\epsilon _{j_1j_2...j_n}(A)_{1k_1}...(A)_{nk_n}(B)_{k_1j_1}...(B)_{k_nj_n}$$

Now I am stuck. But what is frustrating me is that I see exactly what needs to be done for me o get the result. What I am looking for is

$$\sum_{j_i,k_i}\epsilon _{j_1j_2...j_n}\epsilon _{k_1k_2...k_n}(A)_{1k_1}...(A)_{nk_n}(B)_{1j_1}...(B)_{nj_n}$$

So I somehow need to get another epsilon oiver the $k$ terms out of $(B)_{k_1j_1}...(B)_{k_nj_n}$, but I don't know how to motivate this?

Brick
  • 1,345
Meep
  • 3,167
  • This is surely a dullicate of https://math.stackexchange.com/questions/60284/how-to-show-that-detab-deta-detb ! I don't know why some people gave the positive vote! – Arman Malekzadeh May 08 '17 at 16:29
  • 1
    @ArmanMalekzade Hi. As you can see I linked this post in my question. While I am trying to prove the same result, I clearly explained in my question that I am not familiar with the notation used. As you can see I am using very different (at least in my jnderstanding) mathematical objects- the levi civita etc- than what was used. If I am wrong and the answers in the other post do prove the result using the above way, please explain how those answers correspond with my proof attempt. – Meep May 08 '17 at 16:35
  • @ArmanMalekzade The proof above is essentially in alex.jordan's answer on the page linked above. The Levi Civita symbol is essentially the same as the signature of the permutation. – Michael Burr May 08 '17 at 16:47
  • @21joanna12, since the determinant is an invariant of the linear transformation, proofs concerning it are easiest without using coordinates. If transformation $A$ acts on the solid unit cube, then $\det(A)$ is the volume of the resulting solid. I.e.: $\det(A)$ = the coeff. of volume change by $A$.

    Therefore, $\det(AB)$ is the volume of the set that results from acting on the unit cube, first, by $B$, and then by $A$. Multiplicativity follows from this alone.

    For a rigorous proof without coordinates, see the definition of determinant in Halmos's "Finite-dimensional vector spaces."

     – avs May 08 '17 at 16:51
  • @ArmanMalekzade notational differences are not trivial when you're just learning this for the first time. – Ben Grossmann May 08 '17 at 16:51
  • @avs okay, well now you'd need to prove that the determinant you define as the expansion-factor for volume coincides with the usual notion of a determinant. – Ben Grossmann May 08 '17 at 16:52
  • Alex Jordan's proof – Ben Grossmann May 08 '17 at 16:54
  • @MichaelBurr thank you for pointing out Alex Jordans answer. I am trying to see how it relates with mine now. – Meep May 08 '17 at 16:55
  • @Omnomnomnom, the proof expressing the invariant in a particular system of coordinates is much simpler than proving the multiplicativity of determinants in coordinates (i.e., for two matrices rather than two linear transformations). Note the most recent comment from 21Joanna12 as of now.. – avs May 08 '17 at 16:55
  • 1
    @avs sure. But I'd argue that expressing the invariant in coordinates is "the hard part". – Ben Grossmann May 08 '17 at 16:57
  • 1
    @21joanna12 Michael is saying that $$ (-1)^\sigma = \epsilon_{\sigma(1) \sigma(2) \cdots \sigma(n)} $$ in this notation, it's easier to deal with a product of epsilons. – Ben Grossmann May 08 '17 at 17:03

0 Answers0