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Give a counter example to show the following statements are false? $$(A+B)^2=A^2+2AB+B^2.$$

Would it be possible to use the $2\times 2$ matrices to show the statement the above?

Michael Albanese
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shaad
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    That statement is false in general for matrices (or any non-commutative ring). Are you trying to find an example of $2\times 2$ matrices for which the equality fails? – Michael Albanese May 28 '14 at 06:16
  • In general, AB is not equal to BA for matrices. Counter examples are very easy to find. – tpb261 May 28 '14 at 06:19
  • @Michael Albanese By using a simple $2\times 2$ and an identity matrix would be sufficent, wouldn't it? – shaad May 28 '14 at 06:27
  • Identity is of course one of the special cases (it's a unit, it commutes with everything). But taking at random two matrices, it's very unlikely they will commute. Take for instance $(0,1;0,0)$ and $(0,0;1,0)$ (flattened-out notation). – orion May 28 '14 at 06:32
  • Hint. Something like this which is not always true is frequently nearly always false. So if you want a counterexample, don't spend lots of time wondering what it should be: just try something (but avoid "special" cases like the zero matrix and the identity matrix). If your example doesn't do what you want, try another. If the second example doesn't work either it is obviously your unlucky day, give up and go to the next question. OK, that last remark was not completely serious - but it was semi-serious. . . – David May 28 '14 at 06:33

1 Answers1

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Notice that the statement

$$(A+B)^2=A^2+AB+BA+B^2$$

is true. Therefore the statement you want to disprove is equivalent to

$$A^2+AB+BA+B^2=A^2+2AB+B^2$$

i.e.

$$BA=AB$$

It is enough then to find two matrices that don't commute.