$V=\{A\in M_n(\mathbb Q): \operatorname{tr}A=0\}$, 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?

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

• We didn't learned about "linear form", so I'm pretty sure I can't use this solution. – Benny Aug 17 at 20:13
• @Benny "linear form" simply means linear map $V\to W$ such that $\dim W=1$. Have you seen the rank formula(/theorem)? – Arnaud Mortier Aug 17 at 20:15
• 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. – Benny Aug 17 at 20:23
• @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 – Arnaud Mortier Aug 17 at 20:30
• Since we're going to learn it anyway, I think it's not a bad idea to start now. Thanks for your comments :) – 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:

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. $$OE\Delta$$.

• 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 – Arnaud Mortier Aug 21 at 17:23
• @ArnaudMortier: astute of you to point that out. Cheers! – 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.