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Callum Dudley 2022-07-01 Answered
Find the solutions of this linear equation:
( ) X + ( X C ) C = D
where C = ( c 1 , , c n ) R n and D = ( d 1 , , d n ) R n are given and non-zero.
I noticed that the equation ( ) is equivalent to:
( ) C ~ X = D
where C ~ M a t n ( R ) such that: C ~ i , j = { c j c i if  i j 1 + c i 2 if  i = j . To do so, I just said that the i-th component of the LHS of ( ) is ( 1 + c i 2 ) x i + j i x j c j c i
Is there another method to prove that ( ) is equivalent to ( )?
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Answers (1)

Camron Herrera
Answered 2022-07-02 Author has 16 answers
As X and C are real n-vectors,
(1) X + ( X C ) C = X + C ( X C ) (2) = X + C ( X C ) = X + C ( C X ) = ( I + C C ) X .
Explanations:
( 1 ): if u is a vector and s is a scalar, we have [ s ] u = u [ s ], where on the LHS we identify the 1 × 1 matrix [ s ] with the scalar s, so that [ s ] u is the scalar multiplication s u.
( 2 ): the transpose of a real 1 × 1 matrix is the matrix itself.

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