Let u be a strictly positive endomorphism with distinct eigenvalues. Prove that there exists unique v > 0 $v^2 =u$

Question

Let $u \in S(E)$ be a strictly positive endomorphism with distinct eigenvalues. Then prove that there exists unique $v > 0$ s.t. $v^2=u$.

• Can I imagine $u$ as a matrix? Does positive mean that for any vector multiplied with this matrix u the result will be positive?

• How can I use the fact that there are distinct eigenvalues to prove the existence of a unique $v >0$ s.t. $v^2=u$. Can you give me a hint?

Answer 1

I am not sure of some of the terminology but if, as you suggest, we can interpret $U$ as a matrix and if 'strictly positive' means that all the eigenvalues are posiive, then we can write $$U=P \text { diag}(\lambda_1,...,\lambda_n)P^{-1}$$ for an invertible matrix $P$. Then if $$=P \text { diag}(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1},$$ $V^2=U.$

Answer 2

This is my attempt at solving it using our script:

Since u is symmetrical, let's say it has $n*n$ dimensions and following that n eigenvalues say {$l_i$}i=1...n. And ($e_i$)i=1...n is the standard basis

We have that (Proposition 4.17 (1)): u is positive definite (aka $u>0$) if and only if all it's eigenvalues are strictly positive.

Since $u>0 => l_i>0$ for all i and also given from the question $l_i$ != $l_j$ for all i,j with i!=j (not sure why this was included in the question)

And then analogously to the proof of (Proposition 4.17 (4): If u is positive then there exists v ≥ 0 such that $v^2 = u$)

We consider v defined by $v(e_i)$= $(l_i)$^(0.5)*$e_i$. Then the matrix of v in the orthonormal basis e is symmetrical so v belongs to S(E), and we see that v^2 = u and v>0 since $l_i>0$ for all i => ($l_i$)^(0.5)>0 for all i.

I hope you understand my notation, if not just ask.