$V=\{A\in M_n(\mathbb Q): \operatorname{tr}A=0\}$, Prove that $V\oplus \operatorname{Span}\{I_n\}=M_n(\mathbb Q)$ Q: Let $V=\{A\in M_n(\mathbb Q): \operatorname{tr}A=0\}$.
Prove that $V\oplus \operatorname{Span}\{I_n\}=M_n(\mathbb Q)$ 
Since $\dim(V\cap \operatorname{Span}\{I_n\})=0$, $\dim(\operatorname{Span}\{I_n\})=1$ and $\dim(M_n(\mathbb Q))=n^2$, I'm trying (unsuccessfully) to prove that $\dim(V)=n^2-1$, so I can use the dimension thm.
How do I do that?
 A: You are correct to try using what you call the dimension theorem. 
Note that $V$ is the kernel of a non-zero linear form, namely $M_n(\Bbb Q)\to \Bbb Q: M\mapsto m_{1,1}+\ldots +m_{n,n}$. Therefore $V$ has codimension $1$ by the rank-nullity theorem.
A: This holds over any field $\Bbb F$ such that
$0 \not \equiv n \in \Bbb F, \tag 0$
that is, for fields $\Bbb F$ for which
$\text{char}(\Bbb F) \not \mid n; \tag{0.5}$
and can be proved directly from first principles, to wit:
Let
$V_{\Bbb F} = \{A \in M_n(\Bbb F): \operatorname{tr}(A) = 0 \}; \tag 1$
it is clear that $V_{\Bbb F}$ is a vector space over $\Bbb F$, since for
$B, C \in V_{\Bbb F} \tag 2$
we have
$\operatorname{tr}(B + C) = \operatorname{tr}(B) + \operatorname{tr}(C) = 0 + 0 = 0, \tag 3$
and, for
$\alpha \in \Bbb F, \tag 4$
$\operatorname{tr}(\alpha B) = \alpha \operatorname{tr}(B) = \alpha(0) = 0. \tag 5$
Likewise,
$\operatorname{Span}\{I_n \} = \{\alpha I_n, \; \alpha \in \Bbb F \} \tag 6$
is also a vector subspace of $V_{\Bbb F}$, being cleary closed under addition and scalar multiplication.
Now suppose
$B \in M_n(\Bbb F), \tag 7$
and consider 
$B - n^{-1}\operatorname{tr}(B) I_n; \tag 8$
we have
$\operatorname{tr}(B - n^{-1}\operatorname{tr}(B) I_n) = \operatorname{tr}(B) - \operatorname{tr}(n^{-1}\operatorname{tr}(B)I_n)$
$= \operatorname{tr}(B) - n (n^{-1}\operatorname{tr}(B)) = \operatorname{tr}(B) - \operatorname{tr}(B) = 0; \tag 9$
thus
$B - n^{-1}\operatorname{tr}(B) I_n \in V_{\Bbb F}; \tag{10}$
furthermore,
$ B = (B - n^{-1}\operatorname{tr}(B) I_n) + n^{-1}\operatorname{tr}(B) I_n, \tag{11}$
where
$n^{-1}\operatorname{tr}(B) I_n \in \operatorname{Span} \{I_n \}; \tag{12}$
since these assertions hold for arbitrary $B \in M_n(\Bbb F)$ we have shown that
$M_n(\Bbb F) = V_{\Bbb F} + \operatorname{Span}\{I_n \}; \tag{13}$
now if
$C \in V_{\Bbb F} \cap \operatorname{Span}\{I_n \}, \tag{14}$
then
$\operatorname{tr}(C) = 0, \tag{15}$
but
$C \in \operatorname{Span}\{I_n\} \Longrightarrow C = \alpha I_n, \; \alpha \in \Bbb F, \tag{16}$
and
$\operatorname{tr} (C) = n\alpha; \tag{17}$
combining (15) and (17),
$n\alpha = 0; \tag{18}$
by virtue of (0) this yields
$\alpha = 0, \tag{18}$
and thus
$C = 0, \tag{19}$
and so we conclude that
$M_n(\Bbb F) = V_{\Bbb F} \oplus \operatorname{Span} \{I_n \}, \tag{20}$
the requisite result.  $OE\Delta$.
A: $tr$ being a linear form on a $n^2$ dimensional space, its kernel is $n^2-1$ dimensional.
