Dimension of trace-subspace

Question

"Show that V, the set of all square matrices whose trace is 0, form a subspace of $M_{n,n}$ (the set of all square matrices). What is its dimension?"

I have shown that $V$ is a subspace, but I do not know how to determine its dimension yet alone what a base of $V$ would possibly look like. Hints and solutions are appreciated!

Thanks.

Answer 1

A combinatorial proof: you can create a basis for this subspace considering the $n(n-1)$ matrices of the form $(\delta_{ij})$ for $j\neq i$ (which all have trace $0$) together with the $n-1$ matrices of the form $$\begin{pmatrix} 1 & 0 & 0 &\ldots & 0\\ 0 & -1 & 0 & \ldots & 0\\ 0 & 0 & 0 & \ddots & \vdots\\ \vdots & \vdots & \ddots & 0 & 0\\ 0 & 0 & \ldots & 0 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 & 0 &\ldots & 0\\ 0 & 0 & 0 & \ldots & 0\\ 0 & 0 & -1 & \ddots & \vdots\\ \vdots & \vdots & \ddots & 0 & 0\\ 0 & 0 & \ldots & 0 & 0 \end{pmatrix}, ~\ldots ~, \begin{pmatrix} 1 & 0 & 0 &\ldots & 0\\ 0 & 0 & 0 & \ldots & 0\\ 0 & 0 & 0 & \ddots & \vdots\\ \vdots & \vdots & \ddots & 0 & 0\\ 0 & 0 & \ldots & 0 & -1 \end{pmatrix}$$

The dimension of this space is

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

Answer 2

Trace is a nonzero linear function $M_{n,n}\to\Bbb R$, so it has rank$~1$ (since at arrival the space has dimension$1$, it cannot be more than that). Now use the rank-nullity theorem.

Answer 3

Let us note that $\mathrm{tr}: _PM_n(P)\to _PP$ is an epimorphism of vector spaces, so $$ M_N(P)/\mathrm{Ker}tr\cong P. $$ So, we obtain $\mathrm{dim}_P\mathrm{Kertr} = \mathrm{dim}_PM_n(P)-\mathrm{dim}_PP= n^2-1$