Self-adjoint operator (Bens question at Yahoo Answers)

[Fernando Revilla]

Self-adjoint operator (Ben's question at Yahoo! Answers)

Here is the question:

I have matrix that represent T:V -> V (linear map over $\mathbb{R}$) according to basis $E$.

E is not an orthonormal set.

how can I know if this T is self-adjoin ?

I know that if E was orthonormal basis we would take the transpose matrix.

but what about here ?

the matrix is :

1 2
2 1

Here is a link to the question:

Self-adjoint and properties? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Ben,

We need the expression of the inner product. Suppose $G$ is the Gram matrix of the inner product with respect to $E$ and $A$ is the matrix of $T$ with respect to $E$. Denote $X,Y$ the coordinates of $x,y$ respectively with respect to $E$. Then,

$$T\mbox{ is self-adjoint}\Leftrightarrow\; <T(x),y>=<x,T(y)>\quad \forall x\forall y\in V$$
We can express
$$<T(x),y>=(AX)^TGY=X^TA^TGY\\<x,T(y)>=X^TG(AY)=X^TGAY$$ Then, $$X^TA^TGY=X^TGAY\Leftrightarrow X^T(A^TG-GA)Y=0$$ This happens for all $X,Y$ if and only if $A^TG=GA$. So, $$\boxed{T\mbox{ is self-adjoint}\Leftrightarrow A^TG=GA}$$ Particular case: If $E$ is orthonormal then $G=I$, so $T$ is self-adjoint if and only if $A^T=A$ (i.e. $A$ is symmetric).

 

[Ackbach]

$$T\mbox{ is self-adjoint}\Leftrightarrow\; <T(x),y>=<x,T(y)>\quad \forall x\forall y\in V$$

You can use the \langle and \rangle commands in $\LaTeX$ to get better-looking inner products:

$$\langle T(x),y \rangle = \langle x,T(y) \rangle \quad \forall x,y \in V.$$

 

[Fernando Revilla]

You can use the \langle and \rangle commands in $\LaTeX$ to get better-looking inner products:

Thanks, I knew those commads but for me is better-looking < and >. Only a question of particular taste. :)

 

[CellCoree]

Hi Ben,

Thank you for your question. A self-adjoint operator is a linear operator that is equal to its own adjoint. The adjoint of a linear operator T on a finite-dimensional inner product space V is the unique linear operator T* on V such that <T(v), w> = <v, T*(w)> for all v, w in V, where <,> denotes the inner product.

In order to determine if a linear operator is self-adjoint, you can use the following theorem: A linear operator T on a finite-dimensional inner product space V is self-adjoint if and only if its matrix representation with respect to an orthonormal basis is equal to its own transpose.

In your case, since the basis E is not orthonormal, you cannot directly take the transpose of the matrix representation to determine if T is self-adjoint. However, you can still use the above theorem by first finding an orthonormal basis for V and then computing the matrix representation of T with respect to that basis. If the resulting matrix is equal to its own transpose, then T is self-adjoint.

In general, it is important to note that the property of being self-adjoint is dependent on the choice of basis. Therefore, it is always helpful to first find an orthonormal basis for V before determining if a linear operator is self-adjoint.

I hope this helps clarify the concept of self-adjoint operators. Let me know if you have any further questions.

Best, (Scientist)