# Adjoint and inverse of product of operators

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1. Feb 8, 2015

### ellilu11

I know for two linear operators $$H_1, H_2$$ between finite dimensional spaces (matrices) we have the relations (assuming their adjoints/inverses exist):
$$(H_1 H_2)^* = H_2^* H_1^*$$ and $$(H_1 H_2)^{-1} = H_2^{-1} H_1^{-1}$$
but does this extend to operators in infinite dimensions? Thanks.

Last edited: Feb 8, 2015
2. Feb 8, 2015

### Svein

Well, given a dimension N, the relations are true. Since we do not impose any restriction on N...

3. Feb 8, 2015

### Hawkeye18

Yes, this is true for bounded linear operators between Banach spaces.

4. Feb 8, 2015

### ellilu11

Okay thanks. Do you have a reference where I can learn about the Banach space case?

5. Feb 8, 2015

6. Feb 9, 2015

### Fredrik

Staff Emeritus
Is there a definition of the adjoint of a bounded linear operator on a Banach space? The definition I'm familiar with is for Hilbert spaces.

In the finite-dimensional case, the proofs go like this:
\begin{align}
&\langle (AB)^*x,y\rangle =\langle x,ABy\rangle =\langle A^*x,By\rangle =\langle B^*A^*x,y\rangle\\
&B^{-1}A^{-1}A B= B^{-1} I B =B^{-1} B=I.
\end{align} The proofs for bounded linear operators on an infinite-dimensional Hilbert space are the same. For unbounded linear operators, the domains of the operators are an issue. The domains are proper linear subspaces of the Hilbert space, so the first calculation can at best make sense for all $x,y$ in some proper linear subspace (I don't have time to think about the details right now), and the second calculation should be changed to
$$B^{-1}A^{-1}A Bx= B^{-1}I|_{\operatorname{dom} A}Bx =B^{-1}Bx=x,$$ for all x such that x is in the domain of B and Bx is in the domain of A.

Last edited: Feb 9, 2015
7. Feb 9, 2015

### micromass

Staff Emeritus
Yes, although not everybody calls it the adjoint. If you have a operator $T:B_1\rightarrow B_2$, then you can define $T^*:B_2^*\rightarrow B_1^*$ by $T^*(f) = f\circ T$. Only in Hilbert spaces does this give rise to an operator $B_2\rightarrow B_1$ by the Riesz isomorphisms.

8. Feb 9, 2015

### mathwonk

notice from micromass' definition the formula (ToS)* = S*oT* is immediate. i.e. since by definition,

we have T*(f) = foT, applying S* to both sides gives S*oT* (f) = foToS = (ToS)*(f).

so it has nothing to do with the dimension being finite. and i think you can finesse fredrick's subtle remarks in case the maps are not defined everywhere, by saying, "and hence this holds wherever both sides make sense".

Last edited: Feb 9, 2015
9. Feb 9, 2015

### Hawkeye18

If you adapt the pretty standard in functional analysis notation $\langle x, f\rangle$ for $f(x)$, where $x$ is in a Banach space $X$ and $f$ is in its dual (i.e. a bounded linear functional on $X$) then the formula defining the adjoin (dual) operator is absolutely the same as for the Hilbert spaces. Only now, as micromass mentioned the dual operator acts between dual spaces. And the proof Frederik presented for $(AB)^*=B^*A^*$ for Hilbert spaces works in the case of Banach spaces without any changes.