# Trace(matrix) = 0 and the dimension of subspace

## Homework Statement

Claim: Matrices of trace zero for a subspace of M_n (F) of dimension n^2 -1 where M_n (F) is the set of all nxn matrices over some field F.

## Homework Equations

Tr(M_n) = sum of diagonal elements

## The Attempt at a Solution

I view the trace Tr as a linear transformation Tr: M_n (F) -> F. I find the dimension of the subspace spanned by M_n (F) by using the fact:

dim M_n (F) = dim Tr [M_n (F)] + dim null [M_n (F)]

Since dim M_n (F) = n^2, we have

n^2 = dim Tr [M_n (F)] + dim null [M_n (F)]

I'm confused on what's in the right part of the above equation:

Isn't dim Tr [M_n (F)] = 0 since Tr [M_n (F)] = 0?

Shouldn't dim null [M_n (F)] =n^2 - dim Tr [M_n (F)] = n^2 - 1?

Also, I know that the subspace is generated by the following matrices: A, B, and [A,B] (the commutator of A and B).

Thanks!

You said

> Isn't dim Tr [M_n (F)] = 0 since Tr [M_n (F)] = 0?

No, dim Tr [M_n (F)] = 1 since Tr [M_n (F)] = F
You have confused M_n (F) (all nxn matrices over F) with its subset of trace zero matrices. And a (non degenerate) field over itself obviously has dimension 1 because it's spanned by its element 1 (multiplicative identity). So you get indeed

dim null [M_n (F)] =n^2 - dim Tr [M_n (F)] = n^2 - 1