Basis set of matrices with trace 0 can be like that $$\{E_{ij}:1\le i,j\le n \land i\ne j\}\cup\{E_{ii}-E_{i+1,i+1}:1\le i<n\}$$ this set contains $n^2-1$ elements, so space of $n\times n$ matrices with trace 0 is $n^2-1$ dimensional. Space of all matrices $n\times n$ is $n^2$ dimensional and set of matrices with trace equals 0 and not equals 0 are distinct so this should be true: ? $$\dim M = \dim M_{t=0} + \dim M_{t\ne 0}$$ where $M$ - Set of all matrices, $M_{t=0}$ - set of matrices with trace 0 and $M_{t\ne 0}$ - set of matrices with trace not equal 0. $$n^2 = (n^2-1)+(1)$$ So, $$\dim M_{t\ne 0}=1$$ thats correct?
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6$\begingroup$ What makes you think that the set of traceful matrices is a subspace? $\endgroup$– user137731Jan 8, 2016 at 5:33
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$\begingroup$ Right.... Thanks! :) $\endgroup$– TheKwiatek666Jan 8, 2016 at 5:38
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$\begingroup$ What does $\dim X$ mean when $X$ is not a linear or affine space? $\endgroup$– copper.hatJan 8, 2016 at 6:31
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It is not correct. The reason is that the set of matrices with non-zero trace is not a subspace, as can be seen by the simple fact that the zero matrix does not lie in that set. Indeed, if $V$ is any vector space, and $U$ is a subspace of $V$, then $V\setminus U$ is never a subspace of $V$.
However you can do a very similar decomposition:
$$M = M_{t=0} + \mathbb{R}I$$
where $I$ is the identity matrix. Both $M_{t=0}$ and $\mathbb RI$ are subspaces, and they are linearly independent (and even orthogonal under the Hilbert-Schmidt scalar product). Therefore you can write any matrix $A$ uniquely as $$A = X + \lambda I$$ where $X\in M_{t=0}$ and $\lambda$ determines the trace (more exactly, $\operatorname{tr}A = \lambda/\!\dim M$). Clearly, $\mathbb RI$ has dimension 1.
