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 $\{w_1,w_2\}$.

  • $\begingroup$ Also, my question looks very nice rendered using your mathjax, thanks!{Almost symmetric :) } $\endgroup$ Mar 9 '15 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 '15 at 11:42
  • $\begingroup$ Are you still here? $\endgroup$ Mar 12 '15 at 11:04
  • $\begingroup$ @GerryMyerson Yes, and I have made some progress on my ideas here. I will update my context $\endgroup$ Mar 13 '15 at 11:31
  • $\begingroup$ Also yes your concerns above were built on a (correct) representation of my poor understanding!! $\endgroup$ Mar 13 '15 at 11:34

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.