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I have 2 subspaces of $M_2({\bf R})$:

$U =\left\{ \pmatrix{a&b\cr c&d\cr} : c \ge 0 \right\}$

$V = \left\{ \pmatrix{a&b\cr c&d\cr} : c + 2d = 0, a + b - 2c = 0 \right\}$

I need to

1) prove that U isn't a subspace of $M_2({\bf R})$

2) prove that V is a subspace of $M_2({\bf R})$

If someone could guide me for 1), I'll do 2) myself of course.

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What happens when you try to check if $U$ is closed under scalar multiplication? – Ragib Zaman May 7 '12 at 5:02
What is $M_2{(R)}$? $M_2{(R)}$ means the ring of matrixes with elements from the real number? – Babak Miraftab May 7 '12 at 5:02
When you write that you have two subspaces, of course you mean you have two subsets. – Gerry Myerson May 7 '12 at 5:44
up vote 1 down vote accepted

Different approaches work for 1) and 2).

$S$ is a subspace iff whenever $x,y \in S$ and $\lambda \in \mathbb{R}$, then (1) $x+y \in S$, and (2) $\lambda x \in S$.

For $U$, following @babgen, notice that $x=(0,0,1,0) \in U$, but $(-1)x = (0,0,-1,0)$ is not. So it fails (2), hence it is not a subspace.

For $V$, you can work from the definition. For example, choose $(a,b,c,d), (a',b',c',d') \in V$. Since they are in $V$, you know that $c+2d=0$ and $a+b-2c=0$. Similarly, you know that $c'+2d'=0$ and $a'+b'-2c'=0$. Adding together you see that $(c+c')+2(d+d') =0$ and similarly for the other constraint. So it follows that $(a+a',b+b',c+c',d+d') \in V$. So Property (1) of the definition is true. You can prove the other property similarly.

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I feel like I have to emphasise this more (even though both other answers use it). When you see it for the first time it is somewhat amazing that you have: $$\text{A subset }S\text{ is a subspace}\Leftrightarrow S\text{ is closed under addition and scalar multiplication}$$ So what you do when you have exercises like the one above is try to prove conditions $(1)$ and $(2)$ of @copper.hat's answer. If it's possible then it is usually easy (problem 2)) and if it's not possible you will usually find a reason rather quickly (problem 1)). Edit: And to build intuition: Linear equations usually work, inequalities and non-linear equations don't.

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You want to check that $U$ is not a subspace of $M_2(\mathbb{R})$. So you need to verify that one of two things fails: $$\begin{align} A \in U &\Rightarrow \lambda A \in U \text{ for all }\lambda \in \mathbb{R}\\ A, B\in U &\Rightarrow A+B \in U.\end{align} $$ For the first part pick $$ A = \pmatrix{0 & 0\cr 1 & 0\cr} \in U$$ and note that $A$ indeed is an element in $U$. Now we pick $\lambda = -1$. If $U$ is a subspace of $M_2(\mathbb{R})$, then we would have that $\lambda A \in U$, i.e. that $$\lambda A = \pmatrix{0 & 0\cr 1 & 0\cr} = -1 \pmatrix{0 & 0\cr 1 & 0\cr} = \pmatrix{0 & 0\cr -1 & 0\cr}.$$ But this matrix is clearly not in $U$. Hence $U$ is not a subspace of $\mathbb{R}$.

For your second subset $V$, you simply want to check that both of the two conditions are actually satisfied. This will require a bit more work though, and since you said that you want to do this on your own, then I will let you do that. If you want hints, you can ask.

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