# Basis for matrices of trace 0

• September 29th 2010, 07:39 PM
HelloWorld2
Basis for matrices of trace 0
Hey, I was wondering if someone could help me find a basis for the space of all nxn matricies, over R, with trace 0.
• September 29th 2010, 07:53 PM
Quote:

Originally Posted by HelloWorld2
Hey, I was wondering if someone could help me find a basis for the space of all nxn matricies, over R, with trace 0.

It's been a long while, but maybe I can give a small push. Take this with a grain of salt

We know that
$\mathrm{tr}(A) = a_{11} + a_{22} + \dots + a_{nn}=\sum_{i=1}^{n} a_{i i}$

If $\mathrm{tr}(A) = 0$ then all $a_{i i} = 0$.

A basis for the space of all nxn matricies, then, is the set of all matricies with 1 as an ij-th entry **, and all the rest being 0. go through all the possibilities.

ex. for a 2x2 matrix, the basis would be

$
\left[ {\begin{array}{cc}
0 & 1 \\
0 & 0 \\
\end{array} } \right]
$
and $
\left[ {\begin{array}{cc}
0 & 0 \\
1 & 0 \\
\end{array} } \right]
$

EDIT:// ** with $i \neq j$
• September 29th 2010, 08:52 PM
HelloWorld2
But if the 1,1th entry is 1, and the 2,2th entry is -1, we also have a zero trace.
• September 29th 2010, 08:55 PM
Quote:

Originally Posted by HelloWorld2
But if the 1,1th entry is 1, and the 2,2th entry is -1, we also have a zero trace.

oh yeah, sorry. what a silly mistake!

EDIT:// now I'm wondering how you would generalize, b/c I think the form where you have a and -a on the diagonal, replacing the entries for the previous example basis' would be sufficient for 2x2 matricies.
• September 30th 2010, 06:29 AM
HallsofIvy
A basis consists of all n by n matrices in which
1) For i and j from 1 to n with $i\ne j$, $A_{ij}$ with entries $a_{ij}= 1$ and all other entries 0 and
2) For i and j from 1 to n with $j> i$, $B_{ij}$ with entries $b_{ii}= 1$ and $b_{jj}= -1$.
and all other entries 0.
If n= 2, those would be $B_{12}= \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$, $A_{12}= \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$, and $A_{21}= \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}$.