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Examine that $W$ is a subspace of the vector space $M_{3x3}$

$W=({A:det(A) \ge} 0)$

I have two conditions:

1) $ \vec u + \vec v \in W $

2) $ \alpha \cdot \vec u \in W $

$ B,C \in W $

$det(B) + det(C) \ge 0 $

I don't know any properties of determine of matrix which could help with these.

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    I don't understand your attempt: you've just written down four inequalities, with no clue as to what you're doing with them. Please explain further.2012-12-02
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    @ChrisEagle - do you think that the claim is true ? how about $diag(-2,-8,1)+diag(3,3,3)$ as a counterexample ?2012-12-02
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    @ChrisEagle - or take something in $W$ and the scalar as $-1$ then $det$ changes signs2012-12-02
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    Yeah, let $A\in W$. Then $(-A)\in W$. But $A+(-A)=0\notin W$, contradiction, therefore etc. etc. etc.2012-12-02
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    @Belgi: Why are you telling me this? Of course I can solve this problem. Post an answer if you want to.2012-12-02
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    It was wrong what i've wrote. I have no idea how start with it.2012-12-02
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    @ChrisEagle I trust that you know the answer, but because you asked the OP about his effort that threw me off to think that he is trying to prove something that correct...2012-12-02

2 Answers 2

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This claim is not true:

First, note that $A=diag(-2,-8,1),B=diag(3,3,3)\in W$ but $A+B\not\in W$.

Moreover, if $A\in W$ then $|-A|=(-1)^{3}|A|\leq0$ and so $-A\in W$ iff $|A|=0$ but there are matrices in $W$ with non-zero determinant

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$W=\{A\colon \det A\geq 0\}$ is not a subspace of the vector space $M_{3\times3}$ (I assume it is over the reals). Take for example $$ I=\begin{bmatrix} 1 & 0& 0\\ 0&1&0\\ 0&0&1 \end{bmatrix}. $$ Obviously $\det I=1$ and $I\in W$. Now take $(-1)I$ and observe that $\det (-1)I=(-1)^3\det I$.