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?

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    $\begingroup$ What do you mean by $sp\{I_n\}$ ? $\endgroup$ – Jean Marie Aug 17 at 19:54
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    $\begingroup$ the vector space which is spanned by the identity matrix $\endgroup$ – Benny Aug 17 at 19:55

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.

  • $\begingroup$ We didn't learned about "linear form", so I'm pretty sure I can't use this solution. $\endgroup$ – Benny Aug 17 at 20:13
  • $\begingroup$ @Benny "linear form" simply means linear map $V\to W$ such that $\dim W=1$. Have you seen the rank formula(/theorem)? $\endgroup$ – Arnaud Mortier Aug 17 at 20:15
  • $\begingroup$ That's strange to me, because we haven't met linear transformations from a vector space to a field (yet). I think what you mean by the rank thm is what i know as the rank-nullity thm. $\endgroup$ – Benny Aug 17 at 20:23
  • $\begingroup$ @Benny $\Bbb Q$ is not considered as a field here, but as an honest $1$-dimensional vector space over itself. You're right about the rank-nullity theorem, it's been a while since I taught these things in English $\endgroup$ – Arnaud Mortier Aug 17 at 20:30
  • $\begingroup$ Since we're going to learn it anyway, I think it's not a bad idea to start now. Thanks for your comments :) $\endgroup$ – Benny Aug 17 at 20:31

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:


$V_{\Bbb F} = \{A \in M_n(\Bbb F): \operatorname{tr}(A) = 0 \}; \tag 1$

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

$\alpha \in \Bbb F, \tag 4$

$\operatorname{tr}(\alpha B) = \alpha \operatorname{tr}(B) = \alpha(0) = 0. \tag 5$


$\operatorname{Span}\{I_n \} = \{\alpha I_n, \; \alpha \in \Bbb F \} \tag 6$

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

$\operatorname{tr}(B - n^{-1}\operatorname{tr}(B) I_n) = \operatorname{tr}(B) - \operatorname{tr}(n^{-1}\operatorname{tr}(B)I_n)$ $= \operatorname{tr}(B) - n (n^{-1}\operatorname{tr}(B)) = \operatorname{tr}(B) - \operatorname{tr}(B) = 0; \tag 9$


$B - n^{-1}\operatorname{tr}(B) I_n \in V_{\Bbb F}; \tag{10}$


$ B = (B - n^{-1}\operatorname{tr}(B) I_n) + n^{-1}\operatorname{tr}(B) I_n, \tag{11}$


$n^{-1}\operatorname{tr}(B) I_n \in \operatorname{Span} \{I_n \}; \tag{12}$

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

$C \in V_{\Bbb F} \cap \operatorname{Span}\{I_n \}, \tag{14}$


$\operatorname{tr}(C) = 0, \tag{15}$


$C \in \operatorname{Span}\{I_n\} \Longrightarrow C = \alpha I_n, \; \alpha \in \Bbb F, \tag{16}$


$\operatorname{tr} (C) = n\alpha; \tag{17}$

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. $OE\Delta$.

  • 1
    $\begingroup$ For completeness, note that the condition on $\operatorname{Char}\Bbb F$ is necessary, since otherwise $\operatorname{Tr}I_n=0$ and hence $$\operatorname{Span}\{I_n\}\subset V_{\Bbb F}$$ and the hopes for a direct sum fail. Interesting point, +1 $\endgroup$ – Arnaud Mortier Aug 21 at 17:23
  • $\begingroup$ @ArnaudMortier: astute of you to point that out. Cheers! $\endgroup$ – Robert Lewis Aug 21 at 17:24

$tr$ being a linear form on a $n^2$ dimensional space, its kernel is $n^2-1$ dimensional.


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