# Finding the adjoint of a linear operator

• Nov 16th 2011, 08:56 PM
AlexP
Finding the adjoint of a linear operator
I'm trying to find an explicit expression for the adjoint of a particular linear operator. Are there any very general techniques for doing so when we know very little about the setting we're working in? All I have is that $T:V \to V$ is a linear operator and that it does have an adjoint. I know nothing about $V$ (inner product, dimension (possibly infinite), etc.).

The book doesn't give any examples like this, and the only one I've found online uses $T:\mathbb{R}^3 \to \mathbb{R}^2$, so we can work with the inner product formulas explicitly to obtain the adjoint.
• Nov 16th 2011, 10:39 PM
Drexel28
Re: Finding the adjoint of a linear operator
Quote:

Originally Posted by AlexP
I'm trying to find an explicit expression for the adjoint of a particular linear operator. Are there any very general techniques for doing so when we know very little about the setting we're working in? All I have is that $T:V \to V$ is a linear operator and that it does have an adjoint. I know nothing about $V$ (inner product, dimension (possibly infinite), etc.).

The book doesn't give any examples like this, and the only one I've found online uses $T:\mathbb{R}^3 \to \mathbb{R}^2$, so we can work with the inner product formulas explicitly to obtain the adjoint.

There is no general formula that I know of besides the following. If you assume that $V$ is a finite dimensional complex inner product space, and you have some endomorphism $T$ then if $\mathscr{B}$ is an orthonormal basis for $V$ then $[T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast$ where on the left hand side $\ast$ denotes the adjoint of the transformation and on the right it denotes the conjugate transpose. Or, the obvious generlization to non-square matrices.
• Nov 16th 2011, 10:54 PM
FernandoRevilla
Re: Finding the adjoint of a linear operator
Quote:

Originally Posted by Drexel28
If you assume that $V$ is a finite dimensional complex inner product space, and you have some endomorphism $T$ then if $\mathscr{B}$ is an orthonormal basis for $V$ then $[T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast$

More general, if $B$ is a basis of $V$ not necessarily orthonormal, $G$ the Gram's matrix, $A$ the matrix of $T$ and $C$ the matrix of $T^*$ (all these elements with respect to $B$) then, $C=G^{-1}A^*G$ .
• Nov 16th 2011, 10:55 PM
Drexel28
Re: Finding the adjoint of a linear operator
Quote:

Originally Posted by FernandoRevilla
More general, if $B$ is a basis of $V$ not necessarily orthonormal, $G$ the Gram's matrix, $A$ the matrix of $T$ and $C$ the matrix of $T^*$ (all these elements with respect to $B$) then, $C=G^{-1}A^*G$ .

Right, you just change the basis to put it with respect to the orthonormal basis.
• Nov 17th 2011, 07:35 AM
AlexP
Re: Finding the adjoint of a linear operator
Yes, I do know that $[T^*]_\beta=[T]^*_\beta$ for an orthonormal basis $\beta$. But since this only works for finite-dimensional vector spaces, I can't use it here. If I work on it for a while longer and don't come up with anything, I'll post here again and give more details. There's a particular aspect that's making it difficult.
• Nov 18th 2011, 07:31 PM
AlexP
Re: Finding the adjoint of a linear operator
I'm getting nowhere, so here's the actual problem:

"Let $V$ be an inner product space, and let $y, z \in V$. Define $T: V \to V$ by $T(x)= \langle x,y \rangle z$ for all $x \in V$. First prove that $T$ is linear. Then show that $T^*$ exists and find an explicit expression for it."

It's easy to show that $T$ is linear. It's the second part I'm having difficulty with. $T^*$ is the unique linear operator such that $\langle T(x), y \rangle = \langle x, T^*(y) \rangle \quad \forall x,y \in V$. In this case $T(x)= \langle x,y \rangle z$, so we need to show $\langle \langle x, y \rangle z , w \rangle = \langle x, T^*(w) \rangle$. What's throwing me off is that if we can get from the LHS to the RHS via manipulation by the usual useful properties of inner products, I see no way to get x into/out of the inner product and alone on the RHS. I'm probably over-complicating things, as I often seem to do. I would appreciate hints (and as always, only hints).
• Nov 18th 2011, 08:32 PM
Deveno
Re: Finding the adjoint of a linear operator
inner-product spaces are usually defined implicitly over a subfield of $\mathbb{C}$.

so let $T^*(w) = \overline{\langle z,w \rangle}y$

then $\langle x,T^*(w) \rangle = \langle x, \overline{\langle z,w \rangle}y \rangle = \langle z,w \rangle\langle x,y \rangle = \langle x,y \rangle\langle z,w \rangle$

$= \langle\langle x,y \rangle z,w \rangle = \langle T(x),w \rangle$, as desired.
• Nov 19th 2011, 08:12 PM
AlexP
Re: Finding the adjoint of a linear operator
Got it. Thanks.