1. ## Operator invertiblitly

Okay so let $H$ be a Hilbert Space and let $S,T:H\rightarrow H$ be bounded operators. I need to prove that:

$ST$ is invertible if and only if $S$ and $T$ are both invertible.

I have proved roughly that $S$ and $T$ both invertible implies $ST$ invertible. Not sure how to prove the converse though.

2. Originally Posted by ejgmath
Okay so let $H$ be a Hilbert Space and let $S,T:H\rightarrow H$ be bounded operators. I need to prove that:

$ST$ is invertible if and only if $S$ and $T$ are both invertible.

I have proved roughly that $S$ and $T$ both invertible implies $ST$ invertible. Not sure how to prove the converse though.
You won't be able to prove the converse because it is not true.

On the Hilbert space $\ell^2(\mathbb{N},\mathbb{C})$, the unilateral shift is the operator T that takes the sequence $(x_1,x_2,x_3,\ldots)$ to $(0,x_1,x_2,\ldots)$. Its adjoint is the backwards shift operator S that takes $(x_1,x_2,x_3\ldots,)$ to $(x_2,x_3,x_4\ldots,)$ (in other words, it shifts the sequence backwards, with the first coordinate falling off the end). Neither T nor S is invertible (see below), but the product ST is the identity operator, which obviously is invertible.

Notice that if you take the product in the reverse order, TS is not the identity. In fact TS takes $(x_1,x_2,x_3,\ldots)$ to $(0,x_2,x_3,\ldots)$. If you shift forwards and then backwards then you end up where you started from, but if you shift backwards and then forwards you lose the first coordinate and cannot retrieve it.

To see why S and T are not invertible, let $e = (1,0,0,\ldots)\in\ell^2(\mathbb{N},\mathbb{C})$. Then $Se=0$, so S is not invertible. Also, $e$ is not in the range of T, so T is not invertible.

3. Thanks for the response but I think you may have misinterperated the question somewhat. $S$ is not the inverse of $T$ or vica-versa. They are seperate operators. I am pretty sure the question is correct as no one has raised the point that it may not be. However I can see your arguement.

4. Originally Posted by ejgmath
Thanks for the response but I think you may have misinterperated the question somewhat. $S$ is not the inverse of $T$ or vica-versa. They are seperate operators. I am pretty sure the question is correct as no one has raised the point that it may not be. However I can see your arguement.
No, S is not the inverse of T or vice versa. But the fact remains that, in the example I gave, ST is invertible but S and T are not. So the result you are trying to prove is definitely not true.

5. Originally Posted by ejgmath
Okay so let $H$ be a Hilbert Space and let $S,T:H\rightarrow H$ be bounded operators. I need to prove that:

$ST$ is invertible if and only if $S$ and $T$ are both invertible.

I have proved roughly that $S$ and $T$ both invertible implies $ST$ invertible. Not sure how to prove the converse though.

Just use the elementary fact from basic set theory that if $f\circ g$ is bijective then $g$ is one-to -one and $f$ is onto. Since the functions commute we get that both are bijective.