Consider A = $\bigl( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \bigr)$ B = $\bigl( \begin{smallmatrix} s & 0 \\ 0 & t \end{smallmatrix} \bigr)$
as elements of $M_2(C)$ where C is the set of complex numbers and s,t are some real numbers.
Problem. Determine for which s,t we have B $\geq\ A^+$ where $A^+$ represents the positive part of A, where we consider as a function on its spectrum.
Since $M_2(C)$ and C($\sigma(A)$) are homomorphic,
A$\in$$M_2(C)$ corresponds to a function $f$$\in$C($\sigma(A)$)
$\sigma(A)$ = the set of eigenvalues of A = {-1,1}
Since $f$=$f^+$ - $f^-$ where $f^+(x)$=max{$f$($x$),0} and $f^-($x$)$= -min{$f$($x$),0} where $f$ ($x$)=$x$
Thus
$f^+$(-1) = 0
$f^+$(1) = 1
Is this approach correct?
Thank you.
Observe that $x^+=\frac{1}{2}(|x|-x)$. It's much easier to compute $|A|$. Indeed, $A^*A=Id$ so $|A|=\sqrt{Id}=Id$. That gives $A^+=(\begin{smallmatrix} 1/2 & -1/2 \\ -1/2 & 1/2 \end{smallmatrix})$. Thus $B-A^+=(\begin{smallmatrix} s-1/2 & 1/2 \\ 1/2 & t-1/2 \end{smallmatrix})$. You just have to compute the eigenvalues of the precedent matrix.