0
$\begingroup$

I'm trying to find a solution here. When it is given a matrix without any element equal to 0 for example:

enter image description here it is easy to find his commutative matrix but when it is given a matrix with some elements equal to zero for example:

enter image description here I'm not sure how to find his commutative matrix.

Can anybody help me?

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ I mean a matrix B such that AB=BA. $\endgroup$ – Student Oct 12 '13 at 21:23
  • 3
    $\begingroup$ Hmm, $$\begin{pmatrix}1&2\\0&0\end{pmatrix}\cdot \begin{pmatrix} a & b\\c&d\end{pmatrix} = \begin{pmatrix} a+2c & b+2d\\0&0\end{pmatrix};\quad \begin{pmatrix}a&b\\c&d \end{pmatrix}\cdot \begin{pmatrix} 1&2\\0&0\end{pmatrix} = \begin{pmatrix} a & 2a\\c &2c\end{pmatrix}$$ So $c = 0$ and $b+2d = 2a$. $\endgroup$ – Daniel Fischer♦ Oct 12 '13 at 21:29
  • 1
    $\begingroup$ In this case, one says, a matrix that commutes with him, instead of his commutative matrix. :) $\endgroup$ – knsam Oct 12 '13 at 23:04
4
$\begingroup$

Note that if $A$ is a square matrix then $A$ commutes with $I,A,A^2,A^3,\dots$ and so with any polynomial $c_0I+c_1A+c_2A^2+\cdots+a_rA^r$ in $A$. Most of the time, those are the only matrices that commute with $A$.

If $A$ is $n\times n$, then (by Cayley-Hamilton) all powers of $A$ can be expressed in terms of $I,A,A^2,\dots,A^{n-1}$. In particular, for $n=2$, $A$ will commute with all matrices of the form $aI+bA$ and, in most cases, with nothing else.

A complete answer can be given in terms of Jordan canonical form.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

\begin{align} AB &= \overbrace{\left(a + \vec{b}\cdot\vec{\sigma}\right)}^{A}\, \overbrace{\left(a' + \vec{b}'\cdot\vec{\sigma}\right)}^{B} = aa' + \vec{b}\cdot\vec{b}' + \left(a\vec{b}' + a'\vec{b} + {\rm i}\,\vec{b}\times\vec{b}'\right)\cdot\vec{\sigma} \\[3mm] BA &= \left(a' + \vec{b}'\cdot\vec{\sigma}\right) \left(a + \vec{b}\cdot\vec{\sigma}\right) = a'a + \vec{b}'\cdot\vec{b} + \left(a'\vec{b} + a\vec{b}' + {\rm i}\,\vec{b}'\times\vec{b}\right)\cdot\vec{\sigma} \end{align}

$$ AB - BA = 2{\rm i}\,\vec{b}\times\vec{b}' = 0 \quad\Longrightarrow\quad \vec{b}' = \mu\vec{b} $$

Given a $2\times 2$ matrix $A \equiv a + \vec{b}\cdot\vec{\sigma}$, all the matrix of the form $a' + \mu\vec{b}\cdot\vec{\sigma}$ commutes with $A$.

For example: \begin{align} A &= \left(% \begin{array}{cc} 1 & 2 \\ 0 & 0 \end{array}\right) = \left(% \begin{array}{cc} {1 \over 2} & 0 \\ 0 & {1 \over 2} \end{array}\right) + \left(% \begin{array}{cc} {1 \over 2} & 2 \\ 0 & -\,{1 \over 2} \end{array}\right) \\[3mm]&= {1 \over 2} + \left(% \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) + {\rm i}\,\left(% \begin{array}{cc} 0 & -{\rm i} \\ {\rm i} & 0 \end{array}\right) + {1 \over 2} \left(% \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \\[3mm]&= {1 \over 2} + \sigma_{x} + {\rm i}\sigma_{y} + {1 \over 2}\sigma_{z} = {1 \over 2} + \left(1, {\rm i}, {1 \over 2}\right)\cdot\vec{\sigma} \end{align}

Then, $$ \color{#ff0000}{\large B} = \nu + \mu\left(1, {\rm i}, {1 \over 2}\right)\cdot\vec{\sigma} = \nu + \mu\,\left(% \begin{array}{cc} {1 \over 2} & 2 \\ 0 & -\,{1 \over 2} \end{array}\right) = \color{#ff0000}{\large% \left(% \begin{array}{cc} \nu + {1 \over 2}\,\mu & 2\mu \\ 0 & \nu -\,{1 \over 2}\,\mu \end{array}\right)} $$

$\displaystyle{\vec{\sigma}_{i}}$ is a $\tt\mbox{Pauli matrix}$. $\displaystyle{i \equiv x, y, z.\quad}$ $\displaystyle{\vec{\sigma} \equiv \sum_{i = x, y, z}\sigma_{i}\,e_{i}\quad}$ is the $\tt\mbox{Pauli matrix vector}.\quad$ $\displaystyle{e_{x} \equiv \hat{x},\quad e_{y} \equiv \hat{y},\quad e_{z} \equiv \hat{z}}$.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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