Can Someone prove this too?

det(AB)=det(A).det(B)

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- Jan 22nd 2014, 06:26 AM #1

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- Jan 22nd 2014, 09:06 AM #2

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## Re: Det(AB)=Det(A).Det(B)

- Jan 28th 2014, 11:27 AM #3

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## Re: Det(AB)=Det(A).Det(B)

I have a question, I have uploaded a picture of the proof, and I understood everything, however there is something that isn't clear, (it's in Hebrew but I think you can understand what he did.) So in the last part he defined a new permutation sigma from 1 to k1, 2 to k2,..... n to kn, and then he changed the the expression and said that it's equal to detA. However the sign(sigma) (of permutation) is missing? Is there a mistake there? becauseif there is no sign maybe it will be equal to -det(A) at the end? Appreciate it if you can help me out. Thank you.

- Jan 29th 2014, 06:12 PM #4

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## Re: Det(AB)=Det(A).Det(B)

Couldn’t resist opportunity to promote Einstein summation convention to answer your question:

Let A,B,C be matrices, and lAl be detA, etc.

Theorem:

If AB=C then lABl=lCl=lAllBl

Little Preliminaries:

1) lAl=e_{ijk}a_{1i}a_{2j}a_{3k}, definition of determinant

2) e_{ijk}a_{ri}a_{sj}a_{tk}=e_{rst}lAl, Permute rst starting with 123. Any transposition corresponds to a transposition of rows, which changes sign.

Proof:

AB=C, c_{ij}=a_{im}b_{mj}

lABl=e_{ijk}c_{1i}c_{2j}c_{3k}

=e_{ijk}(a_{1r}b_{ri})(a_{2s}b_{sj})(a_{3t}b_{tk})

=(e_{ijk}b_{ri}b_{sj}b_{tk})(a_{1r}a_{2s}a_{3t})

=lBle_{rst}a_{1r}a_{2s}a_{3t}, from 1) and 2)

lABl=lBllAl=lAllBl=lCl

One caveat, think of e_{ijk… }for larger matrices.

note: repeated index is a dummy index: aibi =ajbj=a1b1+a2b2+a3b3+…

also note bonus for determinant of matrix product.

Better than wading through all the summation signs, isn't it?

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