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I need a bit of help with an IFF proof, here it is:

{Let X be a symmetric n × n-matrix. Show: $$X=Y^2$$

for some symmetric matrix Y iff X has only non-negative eigenvalues. }

My thinking:

This requires me to use the principle: $$ A^n=PD^nP^T$$ Now since A is symmetric, any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix. Where P contains the eigenvectors as derived from my eigenvalues in A, $$A=PDP^-1$$ becomes $$A=PDP^T$$

In the >>> direction, if X is Y^2, that means X must be positive definite, so X's eigenvalues will end up being +ve or 0 iff the respective eigenvalue in Y is 0. In the <<< direction I assume all of X's eigenvalues are non-negative, which means if I were to square it, then I just need to show that a matrix times itself is symmetric. ?

Any feedback would be greatly appreciated. Regards

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  • $\begingroup$ "That means X must be positive definite". Why? This is essentially what you need to prove. I don't understand your idea for the other direction. $\endgroup$ – Git Gud Oct 29 '14 at 18:34
  • $\begingroup$ positive definite, so squaring negative values gives positive values, and squaring 0s gives a 0, however that isn't enough to satisfy this proof $\endgroup$ – sahimat Oct 29 '14 at 18:35
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If $Y^2 = X$ for some symmetric $Y$ then write $$ Y = P^TDP \\\implies X = P^TD^2P $$ the eigenvalues of $P^TD^2P$ are the values on the diagonal of $D^2$, that are squares of real numbers. Hence $X\ge 0$ (this is a notation for "$X$ has positive eigenvalues").


If $X$ has positive eigenvalues:

You can write $E = \bigoplus_{i=1}^k E_{a_i}$ where $a_i\ge 0$ and $E_{x} = \{v\in E: Xv = xv \}$.

Define $Yv = \sqrt {a_i} v$ on each $E_{a_i}$, and extend it to $E$ when keeping the linearity. $Y^2 = X$ on each $E_{a_i}$ and also on $E$, using the linearity, and $Y$ is symmetric because $X$ is.

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  • $\begingroup$ I'm currently in class so I'll be giving some more serious consideration over the weekend when I return to my studies. But I won't forget, thank you. Could you clarify your notation, as it's not something I've seen before with the + symbol. $\endgroup$ – sahimat Oct 29 '14 at 19:30
  • $\begingroup$ The symbol just says that you can find a basis of the space made of vectors of one (and only one) of the $E_{a_i}$s. $\endgroup$ – mookid Oct 29 '14 at 20:46

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