# Proof on det(AB) = det(A)*det(B)

1. Oct 22, 2013

### unscientific

1. The problem statement, all variables and given/known data

Let A and B be nxn matrices, show that det(A)=det(A)det(B).

2. Relevant equations

3. The attempt at a solution

I'm not sure how to express each column as a linear combination of vectors purely from A, or even from B. It's more of the dot product form: aT°b

2. Oct 22, 2013

### Erland

3. Oct 22, 2013

### Dick

I think you should tell us what your "general definition of the determinant" is. It's probably easiest to go directly from there.

4. Oct 23, 2013

### unscientific

But i think we should first express the columns of matrix AB in terms of the columns of matrix A? (I'm not sure where B comes in) then somehow show that it is a linear combination, then make use of result 13. Then I think result 14 comes in when you get something that is just jumbled up, which simply gives det (AB) = sgn(P)*det(A)*det(B)

det(A) = Ʃ sgn(P) AP(1)1AP(2)2...

5. Oct 26, 2013

### unscientific

bumpp

6. Oct 26, 2013

### vela

Staff Emeritus
As you noted, the first column of AB is
$$\begin{pmatrix} \sum_{k=1}^n a_{1k}b_{k1} \\ \sum_{k=1}^n a_{2k}b_{k1} \\ \vdots \\ \sum_{k=1}^n a_{nk}b_{k1} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} + \cdots + a_{1n}b_{n1} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} + \cdots + a_{2n}b_{n1} \\ \vdots \\ a_{n1}b_{11} + a_{n2}b_{21} + a_{n3}b_{31} + \cdots + a_{nn}b_{n1} \end{pmatrix}.$$ Express that in the form $c_1\vec{a}_1 + c_2\vec{a}_2 + \cdots + c_n\vec{a}_n$ where $\vec{a}_k$ is the kth column of A. How are the $c_i$'s related to the elements of B?

7. Oct 29, 2013

### unscientific

So, the first column of AB = b11$\vec{a}_1$ + b21$\vec{a}_2$ + .... + bn1$\vec{a}_n$

kth-column = b1k$\vec{a}_1$ + b2k$\vec{a}_2$ + .... + bnk$\vec{a}_n$

I'm just not used to the idea of having "columns within a column", it sort of makes it more than nxn-dimensional.

Also, the earlier problem of

det(A) = λdet(B) + μdet(C)

only had one "special column" in matrix B and C that had a coefficient, while the rest were simply $\vec{a}_1, \vec{a}_2, .... \vec{a}_n$

Last edited: Oct 29, 2013
8. Oct 29, 2013

### vela

Staff Emeritus
Good! So you've done what the first part of the hint suggested, expressing the columns of AB as linear combinations of the columns of A. Now just follow through with the rest of the hint.

9. Oct 29, 2013

### unscientific

Also, the earlier problem of

det(A) = λdet(B) + μdet(C)

only had one "special column" in matrix B and C that had a coefficient, while the rest were simply $\vec{a}_1, \vec{a}_2, .... \vec{a}_n$

while in the new form, it's literally in every column.

10. Oct 29, 2013

### unscientific

I think I got it:

We can simply rearrange the columns of matrix AB so as to separate them into the vectors $\vec{a}_1, \vec{a}_2, \vec{a}_3 ....$ and by result of problem 14, that's not going to change the absolute value of the determinant.

That would solve the dimensional crisis.

Explanation why the product of b's = det(B)

Consider the kth bracket, any permutation within that bracket is equivalent to a permutation of Q(k).