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I need some help, I need to prove the following exercise:

Let A be a definite positive matrix, such that for every positive integer k, there exists a symmetric matrix B such that A=B^k

I haven't thought how to solve it, can anyone give me a hint? Thank you!

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1 Answer 1

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If $A$ is positive definite, its eigenvalues are all strictly positive, and an eigendecomposition exists, $$ A = PDP^{-1}. $$ where $D$ is a diagonal matrix. Then take $$ B = P D^{1/k} P^{-1} $$ so that $A = B^k$.

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  • $\begingroup$ Why do you take D in B as D^1/k? $\endgroup$
    – Ivan Bravo
    Commented May 22, 2020 at 15:59
  • $\begingroup$ Because that is what works. It's a diagonal matrix with the eigenvalues of $A$ raised to the $1/k$ power. When you take $B^2$, for example, you get $(PD^{1/k}P^{-1})(PD^{1/k}P^{-1}) = P D^{2/k} P^{-1}$, and you just keep going until you get back to $A$. $\endgroup$
    – user762914
    Commented May 22, 2020 at 16:00
  • $\begingroup$ Yeah, but, can you say that A is diagonalizable? Why do you guarantee that the eigendecompistion exists? $\endgroup$
    – Ivan Bravo
    Commented May 22, 2020 at 16:14
  • $\begingroup$ Positive definite matrices are invertible, they have all real and positive eigenvalues, they are diagonalizable (the correct definition assumes the matrix is symmetric, some sources drop this and it makes things a mess). These are some of their most basic properties. Go read up on positive definite matrices on wikipedia or snapchat or tiktok or whatever the kids are rocking these days. $\endgroup$
    – user762914
    Commented May 22, 2020 at 16:31

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