problem on set of all matrices whose trace is zero

Let $T(n;\mathbb{R}) \subset M(n;\mathbb{R})$ denote the set of all matrices whose trace is zero. Write down a basis for $T(2;\mathbb{R})$. What is the quotient space $M(n;\mathbb{R})/T(n;\mathbb{R})$ isomorphic to?

Answer of the first part is $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$, $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 1 & -0 \end{pmatrix}$ but i am not sure about second part. My guess is it is $\mathbb{R}-\{0\}$.

-
What is $R$? The reals? –  Julian Kuelshammer Sep 16 '12 at 11:18
@mintu: you should be more precise. 'isomorphic' as vector spaces, groups....?. I don't understand what your answer to the first question means: $[1,0,0,-1] , [0,1,0,0] , [0,0,1,0]$ are matrices?, dimension?. –  Lucien Sep 16 '12 at 11:28
Lucien, I think mintu writes $[a,b,c,d]$ for $\pmatrix{a&b\cr c&d\cr}$. –  Gerry Myerson Sep 16 '12 at 11:55
Ah yes, sorry, I did not see the $2$ in $T(2;R)$. All right. –  Lucien Sep 16 '12 at 12:08

$2^{nd}$ question:
Consider the homomorphism $\tau$: $M(n,\mathbb{R})\to \mathbb{R}$, $A\mapsto tr(A)$.
The kernel is $T(n,\mathbb{R}$) and $im(\tau)=\mathbb{R}$. Then use the first isomorphism theorem (so $M(n,\mathbb{R})/T(n,\mathbb{R})\cong \mathbb{R}$).
Yes you are right. They are linearly independant, and $\dim(T(2;R))=3$ (because $1=\dim M/T=\dim M-\dim T$ and $\dim M=4$) –  Lucien Sep 16 '12 at 12:40