# what is the proof for matrix multiplication being commutative

I understand that if we have matrix $A$ and $B$ then $A \cdot B \neq B \cdot A$ as when you multiply the matrices in a different order, then their cells will shift in another form, thus making their multiplication equate differently in their product matrix.

but what is valid proof i can give to illustrate this

All help is much appreciated

• Do you mean the proof for matrix multiplication being non-commutative? – Omnomnomnom Mar 4 '15 at 22:11

A counter-example would suffice here. Take for example \begin{align*} \pmatrix{1&0\\0&0} \pmatrix{0&1\\0&0} & = \pmatrix{0&1\\0&0}\\ & \qquad\quad\small\diagdown\hspace{-6 pt}||\\ \pmatrix{0&1\\0&0} \pmatrix{1&0\\0&0} & = \pmatrix{1&0\\0&0} \end{align*}
• Does anyone know how to a) remove the manual spacing hack of the $\ne$ and b) rotate said symbol by $90^\circ$? – AlexR Mar 4 '15 at 22:13
• for the spacing you might be able to do something with the align environment. Don't know about rotating the symbol though. – Omnomnomnom Mar 4 '15 at 22:15
• @String Thanks, I tweaked it to \small\diagdown\hspace{-6pt}||: $\small\diagdown\hspace{-6pt}||$ cf. $\displaystyle\ne$ – AlexR Mar 4 '15 at 22:45
• @AlexR: That actually looks quite nice. But to have graphicx among supported packages would be even nicer, though! – String Mar 4 '15 at 22:50
The only proof you need here is an example that shows you that, sometimes, $AB \neq BA$. For example, try $$A = \pmatrix{0 & 1\\0&0}, \quad B = \pmatrix{0&0\\1&0}$$ We find $$AB = \pmatrix{1&0\\0&0}, \quad BA = \pmatrix{0&0\\0&1}$$