k*******y
发帖数: 56 | 1 please forget about what I said, this is the standard proof:
consider the following matrix product
sorry for the messy notation
1
matrix(I,0,-B,xI, nrow=2)*matrix(xI,A,B,xI)
and 2
matrix(xI, -A,0,I) *(xI,A,B,xI)
their det differ by only a factor of a power of x
on the other hand the products are nothing but
matrix(xI,A,0,x^2I-BA)
and
matrix(x^2I-AB,0,B,xI)
now let x=sqrt(\lambda)
can get that the characteristic polymonials of AB and BA
differ only by a vector of \lambda^{m-n}
ie, the nonzero eig | w***t
发帖数: 96 | 2 forget about it.
Just apply elementary operation to
| I A |
| B I |
you will end up with it is equal to |I-AB| as well as |I=BA|.
【在 k*******y 的大作中提到】
: please forget about what I said, this is the standard proof:
: consider the following matrix product
: sorry for the messy notation
: 1
: matrix(I,0,-B,xI, nrow=2)*matrix(xI,A,B,xI)
: and 2
: matrix(xI, -A,0,I) *(xI,A,B,xI)
: their det differ by only a factor of a power of x
: on the other hand the products are nothing but
: matrix(xI,A,0,x^2I-BA)
| d*z
发帖数: 150 | 3 You are right
[ I A ] [ -A I] = [ 0 I ]
[ -B I ] [ I 0] [I+BA -B ]
So
det [I A] = det(I+BA)
[-B I]
同理
det [I -B] = det(I + (-A)(-B))=det(I+AB)
[A I]
而上面左边两个方阵互为转置，所以值相等
即
det(I+BA)=det(I+AB).
【在 w***t 的大作中提到】
: forget about it.
: Just apply elementary operation to
: | I A |
: | B I |
: you will end up with it is equal to |I-AB| as well as |I=BA|.
| b***e
发帖数: 1419 | 4 The trick for this question is use a variable-based radix sort.
In normal radix sort, the bases are fixed. This time use the
variable "n" as the sorting base. Concete soluction is as follows:
Rewrite each number in the representation of (x / n, x % n), where
the "/" represents integeral division, and "%" represents integeral
modulo, as in C. You can first use counting sort to sort the second
elements of all the pairs , and then sort the first element. This
takes O(n) time.
【在 d*z 的大作中提到】
: You are right
: [ I A ] [ -A I] = [ 0 I ]
: [ -B I ] [ I 0] [I+BA -B ]
: So
: det [I A] = det(I+BA)
: [-B I]
: 同理
: det [I -B] = det(I + (-A)(-B))=det(I+AB)
: [A I]
: 而上面左边两个方阵互为转置，所以值相等
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