Tricky matrix inverse

  • Thread starter peterlam
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  • #1
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Suppose A is a invertible n-by-n matrix. Let B be the inverse of A, i.e. B = A^(-1).It is trivial that A = B^(-1).

If we construct a matrix C whose entry is the square of corresponding entry of B, i.e. C_ij = (B_ij)^2, then we compute the inverse of C.

We can compute the inverse of C directly from A without going through the inverse operation twice?

Thank you!
 

Answers and Replies

  • #2
AlephZero
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C is not necessarily invertible, so the answer to your question is "no".

For example
[tex]B = \begin{matrix} 1 & -1 \cr 1 & 1 \end{matrix}[/tex]

[tex]C = \begin{matrix} 1 & 1 \cr 1 & 1 \end{matrix}[/tex]
 
  • #3
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What if we only consider A is positive definite? Then B is positive definite and C should be positive definite too.

Can we compute the inverse of C directly from A in this case?

Thank you!
 
  • #4
AlephZero
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In my counterexample B is positive definite.

x^T B x = x_1^2 + x_2^2

You can write any inverse matrix explicitly in terms of determinants of the matrix and submatrices (this is equivalent to Cramer's rule for solving equations). Think about how a derminant is calculated, and what happens to it if you square all the entries in the matrix. I think it is very unlikely you will get any general result about this.
 

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