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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. – Chris Eagle Dec 2 '12 at 15:43
@ChrisEagle - do you think that the claim is true ? how about $diag(-2,-8,1)+diag(3,3,3)$ as a counterexample ? – Belgi Dec 2 '12 at 15:46
@ChrisEagle - or take something in $W$ and the scalar as $-1$ then $det$ changes signs – Belgi Dec 2 '12 at 15:47
Yeah, let $A\in W$. Then $(-A)\in W$. But $A+(-A)=0\notin W$, contradiction, therefore etc. etc. etc. – Alex Nelson Dec 2 '12 at 15:48
@Belgi: Why are you telling me this? Of course I can solve this problem. Post an answer if you want to. – Chris Eagle Dec 2 '12 at 15:48
up vote 0 down vote accepted

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 $\det (-1)I=(-1)^3\det I$.

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