How to Simplify a Matrix Proof Question - Homework Assistance

[bon]

Homework Statement

Ok so please see question in attachment

Homework Equations

The Attempt at a Solution

Not sure how to solve it..

I'm meant to be using the general defn of determinant as sum over all permutations of sgn(p) A_1P(1)...A_nP(n)

so i see that det A = sum over all permutations of sgn(p) A_1P(1)...A_jP(j)..A_nP(n)

for P(j) = k, but i don't really see how to simplify this into the form they want..

help please :)?

 

[micromass]

Can you show us what you've already tried?

 

[bon]

det A = sum over all permutations of sgn(p) A1P(1)...AjP(j)..AnP(n)

for P(j) = k

I just don't see how to simplify this further :( ..

 

[micromass]

If P(j)=k, then what does A_{j,P(j)} equal in terms of B and C?

 

[bon]

lambda bj + u cj

So i can bring that to the front of the product..

hmm then i can't see how to progress again :(

 

[micromass]

So you have that

A_{1,P(1)}...A_{k,P(k)}...A_{n,P(n)} = \lambda A_{1,P(1)}...B_{k,P(k)}...A_{n,P(n)}+\mu A_{1,P(1)}...C_{k,P(k)}...A_{n,P(n)}

Now this is equal to

B_{1,P(1)}...B_{k,P(k)}...B_{n,P(n)}+\mu C_{1,P(1)}...C_{k,P(k)}...C_{n,P(n)}

Sum this, and you've got what your looking for...

 

[bon]

Thank you very much!

 

[bon]

I've actually decided i still don't understand :(

det A = sum over all permutations of sgn(p) A1P(1)...AjP(j)..AnP(n)

Now AjP(j) = lambda Bjk + u Cjk

so det A = sum over all permutations of sgn(p) A1P(1)...lambda Bjk+u Cjk..AnP(n)

But then how do i sum and simplify?

 

[bon]

So you have that

A_{1,P(1)}...A_{k,P(k)}...A_{n,P(n)} = \lambda A_{1,P(1)}...B_{k,P(k)}...A_{n,P(n)}+\mu A_{1,P(1)}...C_{k,P(k)}...A_{n,P(n)}

Now this is equal to

B_{1,P(1)}...B_{k,P(k)}...B_{n,P(n)}+\mu C_{1,P(1)}...C_{k,P(k)}...C_{n,P(n)}

Sum this, and you've got what your looking for...

sorry, thought id quote you again, so you'd see my response...

 

[micromass]

So what step of my proof don't you get?

Do you see why

A_{1,P(1)}...A_{k,P(k)}...A_{n,P(n)}= \lambdaA_{1,P(1)}...B_{k,P(k)}...A_{n,P(n)}+\mu A_{1,P(1)}...C_{k,P(k)}...A_{n,P(n)}

if P(k)=j??

 

[bon]

oh ok i see.. sorry was being slow/

 

[micromass]

So you get the total proof?

 

[bon]

yes :) thanks!

 

[micromass]

good!

 

[bon]

Now stuck on this one...arghhhh

a hint ? (please see attachment 1 and 2)

 

[micromass]

Do you see why it suffices to show the hint? (every permutation can be written as a product of transpositions)

So it just suffices to prove the theorem for J=(n k). Let's take the case J=(1 2) (thus J is the permutation which switches 1 and 2.

We have to calculate

\det(A^\prime_P)=\sum_{Q}{A^\prime_{1,Q(1)}A^\prime_{2,Q(2)}...A^\prime_{n,Q(n)}}

We know that A^\prime_{j,k}=A_{j,P^{-1}(k)}.

Can you use this in the formula of the determinant?

 

[bon]

I think i get the hint..is it because a permutation that differs by two pair exchanges has sgnP = 1 and three pair exhanges has sgnP = -1 which is the same as one pair exhange..?

Im not sure i do get it actually..

and don't see what J is in your explanation above..

also surely in your expression for det A'p, you've left out sgnQ?

confused :S

 

[micromass]

Yes, how silly of me, I left out sgn(Q). The permutation P is just the one that switches 1 and 2 and leaves the rest fixed. I mean:

P(1)=2, P(2)=1, P(x)=x~\text{if}~1\neq x\neq 2

So, I say you can express A' as

A^\prime_{i,j}=A_{i,P^{-1}(j)}

This is basically the definition of A', do you see that?

Now we have to calculate

\det(A^\prime)=\sum_{Q}{sgn(Q)A^\prime_{1,Q(1)}...A^\prime_{n,Q(n)}}

Can you now change every instance of A^\prime_{j,Q(j)} into an instance of the matrix A?

 

[bon]

Thanks. I'll have a go :)

Just wanted to ask if you could try to explain why the hint works to me clearly..i tried to explain it above but am kind of confusing myself..

sorry to keep bothering you

 

[bon]

So i get det A' = sum over all Q of sgn(Q) A 1 p^-1(Q(1))...A n P^-1(Q(n)).

I guess this is right? where to go from here? :S ahh i hate this topic :P

 

[micromass]

OK, now instead of summing over Q, you can sum over P^{-1}Q. Then you still got every permutation, so you've got (if Q^\prime=P^{-1}Q ):

\sum_{Q^\prime}{sgn(Q)A_{1,Q^\prime(1)}...A_{n,Q^\prime(n)}}

This looks a lot like the formula of the determinant. Only sgn(Q) is bothering us. Can you change Q in a term containing only P and Q'? If you did that, then you just got to use that

sgn(PQ')=sgn(P)sgn(Q')

And then you've got it. This way it doesn't use the hint, so don't worry if you don't understand that...

 

[micromass]

Well, if you really want to know the hint. Do you know what the notation e.g. (1 2)(1 2 3) means?

And don't worry about bothering me, you're not

 

[bon]

Thanks a LOT! I do get it now, but would still like to get the hint... I don't know the notation, no. :(

 

[micromass]

Hmm, explaining the hint is quite troublesome if you don't know that notation.

A transposition is a permutation which simply exchanges two elements. The hint simply says that every permutation can be written as a composition of transpositions. But proving this asks a little work.

 

[bon]

Hmm okay thanks!

The final question now asks me to prove det(AB) = det (A) det (B)

It says that i have to use the previous two results..(which I've asked about)

the hint is to express the comumn vectors of AB as liear combinations of the column vectors of the matrix A..

So i know that (AB)ij = AikBkj but i don't see how to express the columns of AB in terms of column vectors of matrix A

 

[micromass]

Hmmm, that's suddenly a lot harder...

Here's a sketch. (HINT: write this out first in the case n=2, so you see what happens)
First of all let A=(A_1 ... A_n). Check first that

AB=(\sum_{k_1=1}^n{b_{k_1,1}A_{k_1}} \sum_{k_2=1}^n{b_{k_2,2}A_{k_2}} ... \sum_{k_n=1}^n{b_{k_n,n}A_{k_n}} )

Dont worry about the indices k1 to kn, I just wanted other indices for every sum...

So take the determinant from this. You have proved that determinants are multilinear, thus this gives us

\det{AB}=\sum_{k_1,...,k_n=1}^n{b_{k_1,1}...b_{k_n,n}\det(A_{k_1}...A_{k_n})}

Now, a lot of these terms become zero. In fact if any of the k_i equals any of the k_j, then the determinant \det(A_{k_1} ... A_{k_n}) becomes zero. So we can assume that all the k_i are unequal. But then there exists a permutation P, such that P(i)=k_{i}. So we can write the above sum as

\det{AB}=\sum_{P} {b_{P(1),1}... b_{P(n),n}\det(A_{P(1)},...,A_{P(n)})}

Now use the other fact you've proven, and you've got it...

 

[bon]

wow how beautiful. hard though!

Thanks :D

 

[micromass]

Yeah, I find it a bit sadistic that they let you find this on your own...