Find the dimension of the space of $4\times 4$ real matrices with zero trace

Question

I'm wondering if someone can help me to understand this problem.

If $S$ is the subspace of $M_{4,4}(\mathbb{R})$ consisting of all matrices with trace $0$, what is $\dim(S)$?

I've created a matrix with trace $0$, but I can't seem to get the correct answer and I don't understand why.

This is my idea:

$$S = \left[\begin{array}{cccc}0&a&b&c\\d&0&e&f\\g&h&0&i\\j&k&l&0\end{array}\right]$$

Which would have $\dim(S) = 6$. (But this is not correct.)

I would appreciate any help with this.

Thanks,

Rusty

Answer 1

I'm gonna figure that you want 3x3 matrices, though it is easy to extend the answer to 4x4 matrices.

If S = $[a,b,c],[d,e,f],[g,h,i]$ and trace(S) = 0 then by inspection $a+e+i = 0$ so we only have two dimensions because the third diagonal element is determined by the other 2. The other 6 elements can take whatever value they want. So adding it up, it should be 8 dimensions.

EDIT: If you want 4x4 take what I had and extend it (it's easy to do) so you should have 4x4-1 = 15 dimensions.

Answer 2

The trace is a linear form on $\mathcal M_n(\Bbb R)$ that's a linear transformation usually denoted $\operatorname{tr}: \mathcal M_n(\Bbb R)\to \Bbb R$ since we have

$$ \operatorname{tr}(A+\lambda B)=\operatorname{tr}(A)+\lambda \operatorname{tr}(B),\quad \forall A,B\in \mathcal M_n(\Bbb R),\; \forall \lambda\in\Bbb R$$

hence the set of the matrix in $\mathcal M_n(\Bbb R)$ with zero trace is the kernel of this linear form so by the rank-nullity theorem it's a hyper-plane of $\mathcal M_n(\Bbb R)$ i.e. a subspace with dimension $\dim \mathcal M_n(\Bbb R)-1=n^2-1$.

Answer 3

Just to add a different method: let $E_{ij}$ be the elementary matrix with a $1$ in the $(i,j)th$ place and zeroes everywhere else.

Then it is clear that $E_{i,j}$ is a non-zero element of $S$ if $i \neq j$. Also $E_{ii} - E_{nn}$ is a non-zero element of $S$ whenever $i \neq n$. The number of elements counted here is

$$ (n^2 - n) + (n-1) = n^2 - 1. $$

These elements are linearly independent, so are a basis for a subspace of $S$. But $S \lneq M_{n,n}(\mathbb{R})$, so $$ n^2 - 1 \leq \text{dim}(S) < \text{dim}(M_{n,n}(\mathbb{R}) =n^2. $$ So we must have $\text{dim}(S) = n^2 - 1$.

$\textbf{EDIT}$: Sorry, I thought you asked for general $n$, obviously applying the above shows that the dimension of your $S$ is $4^2 - 1 = 15$.