My Question is:

Show that I=$\left\{\begin{pmatrix} a & 0\\ b & 0 \end{pmatrix}\bigg|a,b,c,d \in\mathbb{R}\right\}$

is an ideal of T where T=$\left\{\begin{pmatrix} a & 0\\ b & c \end{pmatrix}\bigg|a,b,c,d \in\mathbb{R}\right\}$

I understand that the definition of an ideal is with a subring has to be closed under addition and if $x\in I$ and $y\in\mathbb{R}$ then $xy\in I$ and $yx\in I$ but I cant seen to apply it to this question.

Any help will be appreciated.

  • $\begingroup$ You should be careful -- in your definition $y$ must be in $T$, not in $\mathbb{R}$. Sometimes people use $R$ for rings which might be the source of your confusion here. $\endgroup$ – walkar Nov 11 '16 at 2:23
  • $\begingroup$ Ah ok thank you for pointing out my mistake @walkar $\endgroup$ – user384716 Nov 11 '16 at 2:25

$\pmatrix{ a & 0\cr b & 0}\pmatrix{u &0\cr v&w}=\pmatrix{au & 0\cr bu & 0}$

$\pmatrix{u &0\cr v&w}\pmatrix{ a & 0\cr b & 0}=\pmatrix{au & 0\cr va+wb &0}$

  • $\begingroup$ Also show the other way. $\endgroup$ – Browning Nov 11 '16 at 2:22
  • $\begingroup$ Have you swapped a,b,c with u,v,w?? $\endgroup$ – user384716 Nov 11 '16 at 2:23
  • $\begingroup$ If you take a random element from I, you can name its entries with a and b. Then when you take a random element from T, you have to use other names, such as u, v, and w. $\endgroup$ – Browning Nov 11 '16 at 2:25
  • $\begingroup$ Ah ok that makes sense. Thanks for pointing it out @Browning $\endgroup$ – user384716 Nov 11 '16 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy