How to Prove det(C) = det(A)det(B)?

[Ylle]

Homework Statement

Let http://imageload.dk/files/370495b6ec270c79fbf66c176f26e442.JPG and

http://imageload.dk/files/6ec41abfcdc8d231593c4a5ee8ee4fe7.JPG

where 0 is the j x k-zeromatrix.

Show that det(C) = det(A)det(B)

Homework Equations

??

The Attempt at a Solution

Got no clue. So I could really use a clue or some help here :)



Regards

 

[VKint]

Hint: Use induction on the dimension of B, and expand the determinant along the bottom row.

 

[Ylle]

Hmmm, not totally sure what you mean.
Induction, isn't that the thing with:
1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2 ?

Not sure how I use that on the dimension of B tbh :?

 

[Count Iblis]

You can also use the definition of a determinant, i.e. the expression involving a summaton over permutations. Then it is trivial.

 

[C.E]

Are you familiar with elementary matrices?

 

[Ylle]

You can also use the definition of a determinant, i.e. the expression involving a summaton over permutations. Then it is trivial.

You mean:
det(A) = a_1jA_1j+...+a_jjA_jj
det(B) = b_1kB_1k+...+b_kkB_kk

And then when I multiply det(A) and det(B), I get the same as if I were to do the determinant of C, which will be det(A)*det(B)-0*0=det(A)*det(B) ?

Or that won't work because they don't have the same dimensions ?

 

[Ylle]

Are you familiar with elementary matrices?

A matrix that does the same to an already existing matrix, as if you were to row-operate ?

 

[Count Iblis]

You mean:
det(A) = a_1jA_1j+...+a_jjA_jj
det(B) = b_1kB_1k+...+b_kkB_kk

And then when I multiply det(A) and det(B), I get the same as if I were to do the determinant of C, which will be det(A)*det(B)-0*0=det(A)*det(B) ?

Or that won't work because they don't have the same dimensions ?

I mean this: If A is an n\times n matrix then, by definition, we have:

\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}

where the summation is over all permutations \pi of the numbers 1\ldots n

 

[Ylle]

I mean this: If A is an n\times n matrix then, by definition, we have:

\det(A) =\sum_{\pi}\operatorname{sign}(\pi)\prod_{i=1}^{n}A_{i,\pi(i)}

where the summation is over all permutations \pi of the numbers 1\ldots n

Ok, that way...

But how does that help me :?
I can make det(A) and det(B) into a linear transformation, but if I do the same on C, won't I just get det(C) = A*B ?

 

[Count Iblis]

Consider some arbitrary permutation \pi. Then consider the contribution to the determinant:

\operatorname{sign}(\pi)C_{1,\pi(1)}C_{2,\pi(2)}\ldots C_{n,\pi(n)}

Where C_{r,s} means the element in the rth row and sth column of C. And n =j+k is the dimension of C.

Now unless the permutation \pi permutes the numbers tranging from 1 to j amongst themselves, you will get zero. So, this means that such permutations \pi can be decomposed as a product of permutations

\pi=\rho\sigma

where \rho permutes the numbers ranging from 1 to j and acts as the identity on the numbers ranging from j+1 to j+k and \sigma is a permutation that permutes the numbers ranging from j+1 to j+k while acting as the identity on the numbers ranging from 1 to j.

Then, if you use that the sign of a product of permutations is the product of the signs of the permutations, you are done.

 

[Ylle]

Hmmm, I'm really sorry. But I'm not sure I understand you here :?
It sounds a bit confusing in my ears - even though it's probably not :S

 

[Count Iblis]

Hmmm, I'm really sorry. But I'm not sure I understand you here :?
It sounds a bit confusing in my ears - even though it's probably not :S

If you are going to choose elements from the first j rows of C, then unless you choose them from submatrix A you'll get zero, right? So, the values you must choose from rows 1 to j are restricted in the range from 1 to j. Therefore the permutation \pi[/tex] will map the range 1 to j to itself.

 

[VKint]

Hmmm, not totally sure what you mean.
Induction, isn't that the thing with:
1 + 2 + 3 + 4 + ... + n = (n)(n+1)/2 ?

Not sure how I use that on the dimension of B tbh :?

Induction is a proof technique that takes advantage of an important property of the integers, namely, well-ordering. It's often compared to pushing down a stack of dominoes. What you do is show that your result holds for some base case (for your exercise, this would be the case where B is a one-by-one matrix), and then show that if it holds for the case corresponding to a natural number n (in your case, B would be an n \times n matrix), then it necessarily holds for n + 1. Carrying this out in your case isn't difficult, if you're familiar with this kind of argument.

However, I actually like Count Iblis's idea better here. Although they're easy to crank out, I try to avoid induction proofs where possible, because they typically don't confer a very intuitive grasp of the theorem upon the reader (or writer). The proof using permutations gives an immediate and intuitive idea of why this result should be true.