Let $(V,\langle\cdot,\cdot\rangle)$ a unitary vector space with $\dim(V)<\infty$ and $g\in L(V,V)$. Define $f:=g \circ g^\text{ad}\in L(V,V)$.

$i)$Show $f$ is self-adjoint.

$ii)$ all eigenvalues of f are non-negative real numbers.

$iii)$ Let $h\in L(V,V)$ self adjoint. Show: If all eigenvalues of h are non-negative, then there exists $k\in L(V,V)$ with $k=k^\text{ad}$ and $h=k\circ k$.

I was able to solve $i)$ and $ii), $ (my definition of scalar product is semi-linear in the first component.)

$$i): \langle f(v),w\rangle=\langle g\circ g^\text{ad}(v),w\rangle=\langle g^\text{ad}(v),g^\text{ad}(w)\rangle=\langle v,g^\text{ad}(g(w))\rangle=\langle v, f(w)\rangle $$

Let $\lambda$ a eigenvalue of $f$ with eigenvector $v$. I use $i)$ in the third equality $$ii):\overline \lambda \langle v,v\rangle=\langle \lambda v,v\rangle=\langle f(v),v\rangle= \langle v,f(v)\rangle=\langle v,\lambda v\rangle=\lambda\langle v,v\rangle$$ Thus $\lambda=\overline \lambda$ and therefore $\lambda$ is real. Suppose $\lambda$ is negative Then $$0>\lambda \langle v,v\rangle=\langle v,f(v)\rangle=\langle v,g(g^\text{ad}(v)\rangle=\langle g^\text{ad}(v),g^\text{ad}(v)\rangle>0$$ Contradiction, thus $\lambda$ is no negative.

I do not know how to start with $iii)$. Some approaches are welcome and a comment about $i)$ or $ii)$ is also welcome!

  • $\begingroup$ Do you know that $V$ has basis consisting of eigenvectors of $h$, in fact an orthonormal basis? $\endgroup$ – fredgoodman Mar 30 '18 at 19:38
  • $\begingroup$ you mean since h is self-adjoint. Its transformation matrix is a diagonalizable? but how to proceed? could you lead me the way? $\endgroup$ – user519338 Mar 30 '18 at 20:14
  • $\begingroup$ You can work in the basis of eigenvectors. The matrix of the linear transformation $h$ in that basis is diagonal with non-negative entries. Can you find a square root of a diagonal matrix with non-negative entries? If so, you will have found a square root of $h$. $\endgroup$ – fredgoodman Mar 30 '18 at 21:57

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