stochastic matrices properties

1 The question is actually easy if you write out all things
Let x=(x_1, x_2,...,x_n)^T
A=
a_11,a_12,...a_1n
a_21,a_22,...a_2n
...
a_n1,a_n2,...a_nn
where x_1+x_2+...+x_n=1 ; a_11+a_21+...+a_n1=1, a_1n+a_2n+...+a_nn=1
Ax=
a_11x_1+a_12x_2+...+a_1nx_n
a_21x_1+a_22x_2+...+a_2nx_n
...
a_n1x_1+a_n2x_2+...+a_nnx_n
Add them altogether
(a_11x_1+a_12x_2+...+a_1nx_n)+(a_21x_1+a_22x_2+...+a_2nx_n)+...+(a_n1x_1+a_n2x_2+...+a_nnx_n)
=(a_11+a_21+...+a_n1)x_1+(a_12+a_22+...+a_n2)x_2+...+(a_1n+a_2n+...+a_nn)x_n
=x_1+x_2+...+x_n
=1
2
The statement is wrong. I can let A and C are abitrary matrices an B is an diagonal matrix. Then
A and B commute, B and C commute
But generally, AC=CA is wrong


2009-02-01 19:21:37 補充：
MR學問﹐我承認看錯題。不過其實這些是雞同蛋的問題﹐轉過形式便PROVE到﹐不至於COMPLETELY WRONG 嘛

2009-02-01 19:26:10 補充：
(a_11+a_21+...+a_n1)x_1+(a_12+a_22+...+a_n2)x_2+...+(a_1n+a_2n+...+a_nn)x_n

=1 (GIVEN)

x_1+x_2+...+x_n=1 (GIVEN)

2009-02-01 19:26:16 補充：
SUBSTRACT

(a_11+a_21+...+a_n1-1)x_1+(a_12+a_22+...+a_n2-1)x_2+...+(a_1n+a_2n+...+a_nn-1)x_n=0

As x_1,x_2,...x_n are positive
=>a_11+a_21+...+a_n1-1=0,...a_1n+a_2n+...+a_nn-1=0

So A , a square matrix of order n, is a stochastic matrix.

You are completely wrong!
Why on earth do you show the converse of the proposition!

2009-02-01 15:59:05 補充：
1. Denote entry at i-th row and j-th colm of matrice A be A(i,j)
Let M = A*x, then, M(i) = SUM[j]{ A(i,j)*x(j) }
Suppose M is stochastic, SUM[i]{ M(i) } = 1,
that is, SUM[i]{ SUM[j]{ A(i,j)*x(j) } } = 1,
grouping x(j) terms and factorizing gives
SUM[j]{ SUM[i]{ A(i,j) }*x(j) } = 1;
on the other hand, as x itself is stochastic,
SUM[j]{ x(j) } = 1.
As all x(j) are positive, the weights 1,1,1,1...,1 are unique!
Thus, SUM[i]{ A(i,j) } = 1,
Therefore, A itself is also stochastic.
.
2. Take A = [ 1 1 ; 2 2 ], B = I , C = [ 2 2 ; 1 1 ]
A*B and B*C are comm trivially,
but AC = [ 3 3 ; 6 6 ] =\= CA = [ 6 6 ; 3 3 ]
Thus the statement is false.
.
Remarks: Note that stochastic matrix is also called transition matrix,
and is the matrix representation of markov chain.
In fact, for any stochastic M, M^n is also stochastic for all natural number n.

2009-02-01 16:00:44 補充：
oops, we should point out that:
all A(i,j)>=0

2009-02-01 16:07:19 補充：
The proof of this point is quite abstract.
M(i) = SUM[j]{ A(i,j)*x(j) } >= 0 as M is stochastic
As our matrix A is fixed and x is arbitrary stochastic column vector,
taking x = [1,0,..,0] gives A(i,1) >= 0 for all i
taking x = [0,1,0,...,0] gives A(i,2) >= 0 for all i
...
Similarly, it holds for all i,j,

2009-02-01 16:09:15 補充：
This is actually the key of attempting proof --
"we have to check that ALL the conditions are satisfied"!