Let H be the subset of $M_2(\mathbb{R})$ consisting of all matrices of the form

$H^* = \left \{ \begin{pmatrix} a &-b \\ b&a \end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$

Is (H*, .) a group under multiplication?

I said no because since $b \neq 0$, the identity matrix can't exist. But my prof briefly said "either a or b can't be 0, not a and b"

Am I right or his wrong?

  • 1
    $\begingroup$ Note that your last line is false (and syntactically incorrect). You're not right and he's not wrong. However you are wrong, and he is right. $\endgroup$ – Asaf Karagila Oct 18 '12 at 1:16

He’s right, and you’re wrong. The definition

$$H^* = \left \{ \begin{pmatrix} a &-b \\ b&a \end{pmatrix} : a,b\in\mathbb{R} , a\neq 0 \; \text{or} \; b \neq 0\right\}$$

makes $H^*$ the collection of all $2\times 2$ real matrices such that at least one of $a$ and $b$ is non-zero. The identity matrix is such a matrix: for it you have $a\ne 0$.

The condition $$a\ne 0\text{ or }b\ne 0$$ is equivalent to the condition $$(a\ne 0\text{ and }b\ne 0)\text{ or }(a\ne 0)\text{ and }b=0)\text{ or }(a=0\text{ and }b\ne 0)\;;$$ the only case that it excludes is $a=b=0$. In mathematics and logic, or is inclusive: ‘$a\text{ or }b$’ is ‘$a\text{ or }b\text{ or both}$’.

  • $\begingroup$ So what is an example of what i mean? That is when $b \neq 0$ ever? How do I say that? $\endgroup$ – Hawk Oct 21 '12 at 5:32
  • $\begingroup$ @jak: If the definition had been $$H^*=\left\{\pmatrix{a&-b\\b&a}:a,b\in\Bbb R,a\neq 0\text{ and }b\neq 0\right\}\;,$$ your objection would have been correct. That definition requires that $a$ and $b$ both be non-zero and therefore excludes the identity matrix. $\endgroup$ – Brian M. Scott Oct 21 '12 at 10:44

In mathematics the use of the word or is inclusive, not exclusive. It is possible that only $b\neq 0$, it is possible that only $a\neq 0$, and it is possible that both $a\neq0$ and $b\neq 0$.

It is only impossible that both $a=b=0$.

So the identity matrix is in $H^\ast$, since $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ has the wanted property.


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