# dimensions and basis

• Nov 15th 2006, 01:00 PM
mesterpa
dimensions and basis
M =
{( a b)
{(-b c) :a,b,c real numbers }

N =
{( x 0)
{(y 0) :x,y real numbers }

are subspaces of M2(R).

I need to find the dimension of M and N. how do i do this
is the dimension the number of elements in the basis?
if so how do i find a basis for these?
thank you
• Nov 15th 2006, 01:42 PM
Plato
The basis for M is:
$\left\{ {\left[ {\begin{array}{rl}
1 & 0 \\
0 & 0 \\
\end{array}} \right],\left[ {\begin{array}{rl}
0 & 1 \\
{ - 1} & 0 \\
\end{array}} \right],\left[ {\begin{array}{rl}
0 & 0 \\
0 & 1 \\
\end{array}} \right]} \right\}.$
• Nov 15th 2006, 02:23 PM
mesterpa
yeah i thought it was but i worked it out just by looking at it. if im doing a proof is there a way i can devise a basis or does it have to satisfy anything to be a basis?
this means the dimension of M = 3?
what is M + N?
• Nov 15th 2006, 03:04 PM
Plato
The sum M+N produces $M_2 (R)$.
To see this consider:
$\left[ {\begin{array}{rr}
x & y \\
w & z \\
\end{array}} \right] = x\left[ {\begin{array}{rr}
1 & 0 \\
0 & 0 \\
\end{array}} \right] + y\left[ {\begin{array}{rr}
0 & 1 \\
{ - 1} & 0 \\
\end{array}} \right] + z\left[ {\begin{array}{rr}
0 & 0 \\
0 & 1 \\
\end{array}} \right] + \left( {y + w} \right)\left[ {\begin{array}{rr}
0 & 0 \\
1 & 0 \\
\end{array}} \right].$