Determinant of a matrix with identity blocks

[penguin007]

Hi all,
I'm studying my mathematics lesson, and there is an example I can't understand:

Consider the matrix
A=(0 In)
(-In 0)
with In the identity nxn
We want to compute detA :

We introduce the permutation
p=(1 2 ... n n+1 ... 2n)
(n n+1 ... 2n 1 ... n )


s.t if we apply p to the columns of A, then we get
B=(In 0)
(0 -In)
(so far so good)

then we say: detB=(-1)^n (ok)
the sign of p is (-1)^n (all right)

then detA=(-1)^n*(-1)^n=1...(why?)


Thanks in advance

 

[tiny-tim]

… then we say: detB=(-1)^n (ok)
the sign of p is (-1)^n (all right)

then detA=(-1)^n*(-1)^n=1...(why?)

Hi penguin007!

PA = B, so detP*detA = detB

 

[vela]

How'd you get the sign of p is $(-1)^n$? That's where the problem lies.

 

[penguin007]

Hi tiny-tim:
I think I got it but I thought a map could be defined by a matrix only if it was linear?? Yet I can see p has a matrix (the one that reverses the columns 1 and n...) so:
is p a linear map? or can some non linear maps have a matrix??

vela:
s=The sign of p is given by (-1)^(the number of inversions by p) I counted n-2 inversions then s=(-1)^n...


Thanks again

 

[tiny-tim]

Hi penguin007!

Hi tiny-tim:
I think I got it but I thought a map could be defined by a matrix only if it was linear?? Yet I can see p has a matrix (the one that reverses the columns 1 and n...) so:
is p a linear map? or can some non linear maps have a matrix??

P is linear …

what makes you think it isn't? ​

 

[vela]

vela:
s=The sign of p is given by (-1)^(the number of inversions by p) I counted n-2 inversions then s=(-1)^n...

Oy, I misread what you wrote in this thread too. I think I'll just shut up now.

 

[penguin007]

tiny-tim:
please correct me: p is a permutation; p is linear would mean, for instance:p(3)=p(1+2)=p(1)+p(2), but p(3)=n+2<>p(1)+p(2)...( I know there is a snag, but where??)

vela:
That's all right, these things happen to everyone.

 

[tiny-tim]

p is a permutation; p is linear would mean, for instance:p(3)=p(1+2)=p(1)+p(2), but p(3)=n+2<>p(1)+p(2)...

oh i see …

no, these Ps acts on n-tuples (or vectors).

So eg P₁₂(a,b,c,d,e,f) = (b,a,c,d,e,f),

and you can check that P₁₂(a+A,b,c,d,e,f) = (b,a,c,d,e,f) + (b,A,c,d,e,f) = P₁₂(a,b,c,d,e,f) + P₁₂(A,b,c,d,e,f)

 

[penguin007]

Thanks a lot tiny-tim for all these explanations, now I understand!