Showing det(C) = det(A)det(B)

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

 

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

 

Are you familiar with elementary matrices?

 

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

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

where the summation is over all permutations [itex]\pi[/itex] of the numbers [itex]1\ldots n[/itex]

 

Consider some arbitrary permutation [itex]\pi[/itex]. Then consider the contribution to the determinant:

[tex]\operatorname{sign}(\pi)C_{1,\pi(1)}C_{2,\pi(2)}\ldots C_{n,\pi(n)}[/tex]

Where [itex]C_{r,s}[/itex] 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 [itex]\pi[/itex] permutes the numbers tranging from 1 to j amongst themselves, you will get zero. So, this means that such permutations [itex]\pi[/itex] can be decomposed as a product of permutations

[tex]\pi=\rho\sigma[/tex]

where [itex]\rho[/itex] permutes the numbers ranging from 1 to j and acts as the identity on the numbers ranging from j+1 to j+k and [itex]\sigma[/itex] 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.

 

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 [itex]\pi[/tex] will map the range 1 to j to itself.

 

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 [tex] B [/tex] is a one-by-one matrix), and then show that if it holds for the case corresponding to a natural number [tex] n [/tex] (in your case, [tex] B [/tex] would be an [tex] n \times n [/tex] matrix), then it necessarily holds for [tex] n + 1 [/tex]. 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.