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I'm preparing for an exam, and this is an exercise that was given to a teammate. It would be helpful for me to solve it, but I'm stuck. This is the statement.

Given $A=(a_{ij}), B=(b_{ij}) $ matrices in $ M_2(\mathbb{K})$ such that $A$ is symmetric, and $B$ is skew-symmetric. Set the conditions to make them commutative with respect to multiplication.

Please note that the course is pretty basic and supposedly should be resolved without having notion of eigenvectors or other concepts of linear algebra, only elementary matrix theory. I solved by writing the matrices and making the operations, which allowed me to make a system of equations and then led me to the following answer:

$A= \left( \begin{matrix} a_{11} & a_{12} \\ a_{12} & -a_{11}\\ \end{matrix} \right) $ ; $B= \left( \begin{matrix} 0 & b_{12} \\ -b_{12} & 0\\ \end{matrix} \right) $

Anyway, I would like to write the deduction in a more formal way, and this is where I get stuck.

This is what I've done so far.

$$ A=(a_{ij}) \mid a_{ij}=a_{ji} \qquad \forall i,j\in \{1,2\} \tag{1} \\$$ $$\\ B=(b_{ij}) \mid b_{ij}=-b_{ji} \qquad \forall i,j\in \{1,2\} \tag{2} $$ Therefore,

$$\ AB=(c_{ij}) \ | \ c_{ij}=\sum_{k=1}^2 a_{ik}b_{kj} \stackrel{(1)}= -\sum_{k=1}^2 b_{jk}a_{ki} \\ BA=(d_{ij}) \ | \ d_{ij}=\sum_{k=1}^2 b_{ik}a_{kj} \stackrel{(2)}= -\sum_{k=1}^2 a_{jk}b_{ki}$$

Then, $$AB=BA \\ \sum_{k=1}^2 a_{ki}b_{jk}=\sum_{k=1}^2 a_{jk}b_{ki}$$

If $i=j \implies b_{ii}=-b_{ii} \implies b_{ii}=0 \quad\forall i\in \{1,2\} $

So far I've come. I do not know if I'm good.

user1551
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Карпатський
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1 Answers1

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Since $B$ is skew-symmetric, you already know $B$ has the form that you derived. So all you have to do is write $AB = BA$ in terms of the matrix entries where $A$ has off-diagonal elements equal and that will give you your equations to solve. Summation notation can make things messy when it's not needed. It's probably easier to let $b$ be the upper right entry of $B$ (which uniquely defines $B$, the diagonal is 0 and the lower left entry is $-b$) and let $a_1,a_2$ be the diagonal entries of $A$ and $a_3$ be the two equal off-diagonal entries of $A$.

user2566092
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  • Thank you for your suggestions. That is actually how I solved the problem the first time. I'm using summation notation because I think it would give me a better understanding if I had to prove it on square matrices of nxn. – Карпатський Feb 09 '15 at 22:14
  • @Andrés I hear what you're saying but I'm pretty sure the problem would be much, much harder for larger matrices. – user2566092 Feb 09 '15 at 22:19