MHB Self-adjoint operator (Bens question at Yahoo Answers)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Operator
AI Thread Summary
To determine if the linear map T represented by the matrix is self-adjoint when the basis E is not orthonormal, one must consider the Gram matrix G associated with the inner product defined by E. The condition for T to be self-adjoint is that the equation A^T G = G A holds true, where A is the matrix of T. If E were orthonormal, the condition simplifies to A being symmetric, or A^T = A. The discussion emphasizes the importance of the inner product's expression and the role of the Gram matrix in establishing self-adjointness. Understanding these concepts is crucial for analyzing linear maps in non-orthonormal bases.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
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.
 
Last edited:
Mathematics news on Phys.org
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).
 
Fernando Revilla said:
$$T\mbox{ is self-adjoint}\Leftrightarrow\; <T(x),y>=<x,T(y)>\quad \forall x\forall y\in V$$

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

$$\langle T(x),y \rangle = \langle x,T(y) \rangle \quad \forall x,y \in V.$$
 
Ackbach said:
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. :)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Back
Top