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A is a square matrix, I denote transpose A as A^T
The following statement is false:
"If AA is symmetric, then so is A. "
The counterexample is A= ((0 1) (0 0)), ie AA=((0 0) (0 0))

However, how to derive a counterexample for this question systemmatically?

2006-09-28 20:38:59 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

it is very hard to find counterexamples...
and there is no method
but in this case:
you can simply try to solve the problem for 2x2 matrices,
so you get:

a b
c d
its square is:
a^2 +bc ... ab+bd
ac+dc ...... cb+d^2
so you assume that the square is symmetric, then
ab+bd =ac+cd
b(a+d)=c(a+d),
so if a+d is not cero, that means that b=c and therefore your original matrix is symmetric,
on the other hand if a+d =0 then the equation is satisfied, and you do not need any conditions on b and c,
so, another counter example is:
0 ..1
-1 .0
one more:
1....-1
1....-1

2006-09-29 06:34:57 · answer #1 · answered by Anonymous · 0 0

Well, the zero matrix is symmetric, so simply use any nonsymmetric matrix for which im(A) ⊆ ker(A). It's kind of pointless though, as you only need one counterexample, and you already have one.

2006-09-29 03:54:49 · answer #2 · answered by Pascal 7 · 0 0

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