Solve a-x-a +x and a+x-a-x
Answers
Answer:
Given AA is d×dd×d and real-valued:
The formula X=AXAX=AXA can be rewritten with XX reordered as vec(X)vec(X), a column vector of d2d2 elements:
vec(X)=(AT⊗A)vec(X)
vec(X)=(AT⊗A)vec(X)
... where ⊗⊗ is the Kronecker product, a d2×d2d2×d2 matrix. So, there are non-zero solutions iff (AT⊗A)(AT⊗A) has any eigenvalues equal to 1, and the solution(s) are the corresponding eigenvectors.
The d2d2 eigenvalues of (AT⊗A)(AT⊗A) can be obtained by multiplying all pairings of the dd eigenvalues of AA. So if any of the eigenvalues of AA have an absolute value of 1 (including complex values), or if any pair of them satisfies λjλk=1λjλk=1, you have a solution to X=AXAX=AXA.
[In your case A=BTBA=BTB is symmetric so all the eigenvalues of AA will be real.]
Of course, it may not be possible to scale it to that XX is a similarity matrix. Clearly, all solutions XX will be singular, unless the determinant of AA is 1 or -1. In fact, any XX obtained directly from an eigenvector of the Kronecker product will be of rank 1, being the outer product of eigenvectors of AA (this is discussed more below), so a non-singular XX solution could be obtained only via a sum of multiple such solutions.
Answer:
0 is the answer
a-x-a+a=0
a+x-a-x=0
Step-by-step explanation:
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