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I want to determine what it means for a symmetric matrix to be written in terms of linear operators.


Perhaps the key? it looks like the form I am looking for might be the self-adjoint operators?


My guess is that it would look something like this:

$$\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} M(T,\{v_1,v_2\},\{w_1,w_2\})=\begin{bmatrix} \phi(v_1)\\ \phi(v_2)\end{bmatrix}=\begin{bmatrix} a&b\\b&c \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$$

Where we have $\phi:V\to V$ from basis $\{v_1,v_2\}$ to $\{w_1,w_2\}$ and this is for the $2\times 2$ matrix case.

For the $3\times 3$ case, we would merely have more operators:

$$\phi(v_1) = av_1 + b v_2 + cv_3$$ $$\phi(v_2) = bv_1 + dv_2 + ev _3$$ $$\phi(v_3) = cv_1 + ev_2 + fv_3$$


Is this a good interpretation?

Where $M(T,\{v_1,v_2\},\{w_1,w_2\}$) refers to the matrix of transformation $T$, taking us from basis $\{v_1,v_2\}$ to $\{w_1,w_2\}$.

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  • $\begingroup$ Also, my question looks very nice rendered using your mathjax, thanks!{Almost symmetric :) } $\endgroup$ Mar 9, 2015 at 19:56
  • $\begingroup$ You write that you are going from one basis to a different one, but you express the result of applying $\phi$ to the elements of one basis in terms of the same basis. Where does the second basis come into it? Usually, when we deal with operators on a vector space, we use the one basis for both argument and value. $\endgroup$ Mar 10, 2015 at 11:42
  • $\begingroup$ Are you still here? $\endgroup$ Mar 12, 2015 at 11:04
  • $\begingroup$ @GerryMyerson Yes, and I have made some progress on my ideas here. I will update my context $\endgroup$ Mar 13, 2015 at 11:31
  • $\begingroup$ Also yes your concerns above were built on a (correct) representation of my poor understanding!! $\endgroup$ Mar 13, 2015 at 11:34

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I don't really understand the way you composed this post, so this might be guesswork from my part.

Do you wish to know that if a linear operator's matrix in some basis is symmetric, then what does it imply of the operator's properties?

Normally, nothing. For a linear operator, symmetry of the matrix is not a property that is preserved by basis change.

If $A:V\times V\rightarrow \mathbb{R}$ is a bilinear functional, and it is symmetric, eg. $$ A(x,y)=A(y,x)\ \ \forall x,y\in V, $$ then its matrix in any basis is gonna be symmetric, since if $\{e_1,...,e_n\}$ is a basis of $V$ then $$ a_{ij}=A(e_i,e_j)=A(e_j,e_i)=a_{ji}, $$ however, this is for bilinear functionals.

For linear operators, however, if $(V,\langle,\rangle)$ is a (real) inner product space, then if $A:V\rightarrow V$ is a linear operator, then you can make it into a bilinear functional $\tilde{A}:V\times V\rightarrow\mathbb{R}$, defined by $$ \tilde{A}(x,y)=\langle x,Ay\rangle. $$ If $A$ is self-adjoint with respect to the inner product, then $\langle x,Ay\rangle=\langle Ax,y\rangle$, and thus $\tilde{A}$ is symmetric. However, the matrix of $A$ and $\tilde{A}$ will only match, if you take the matrix w.r.t an orthonormal basis.

Long story short, if you have an inner product, and your operator has a symmetric matrix with respect to an orthonormal basis, then your operator is self-adjoint. Otherwise, it is just a coincidence and means nothing.

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