Let $A$ be a $C^\ast$-algebra. Is it possible to prove that if $a \ge 0$ then
$ab, ba \ge 0$ if and only if $b \ge 0$?
Asked
Active
Viewed 73 times
2
-
Here are a couple of posts that are relevant for similar spectral theory questions: http://math.stackexchange.com/questions/874324/on-the-spectrum-of-a-product-in-a-banach-algebra-in-specific-case and http://math.stackexchange.com/questions/19576/spectrum-of-a-product-of-operators-on-a-banach-space – Cameron Williams Dec 01 '14 at 02:42
2 Answers
0
No, it is not possible to prove that this is the case.
(non-zero) Counterexample in $\Bbb C^{2 \times 2}$: $$ a = \pmatrix{1&0\\0&0}, \quad b = \pmatrix{1&0\\0&-1} $$
Counterexample to the comment: $$ a = \pmatrix{1&0\\0&0}, \quad b = \pmatrix{1&1\\1&1} $$ note that neither $ab$ nor $ba$ is self adjoint. Note that $ab$ will be self-adjoint iff $a$ and $b$ commute.
Ben Grossmann
- 225,327
-
-
I don't quite understand what you mean here. In any case, notice that for this example $ab = ba = a \neq 0$. – Ben Grossmann Dec 01 '14 at 02:42
-
-
This is certainly not true. Note that $ab,ba$ need not be self-adjoint. Should we add that as an assumption? – Ben Grossmann Dec 01 '14 at 02:49
-
-
It is indeed true (for matrices, not sure about the general case) that if $a,b$ commute with $a,b\geq 0$, then $ab \geq 0$. – Ben Grossmann Dec 01 '14 at 04:10
0
Simplest counterexample: $a=0$.
Somewhat less trivial: in $C([0,1])$ let $a$ be a function that is $0$ on $[0,1/2]$ and nonnegative. Then $ab = ba \ge 0$ if $b \ge 0$ on $[1/2, 1]$, but you could have $b(t) < 0$ somewhere in $[0,1/2)$.
Robert Israel
- 448,999