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i) $2x_1 - 3x_2 = 5$

I know that this is not a linear subspace of $\mathbb{R^2}$ since $2(0)-3(0)=0 \neq 5$. But for

ii) $3x_1 + 4x_2 = 0$

do I just need to find values for $x_1$ and $x_2$ that work for the equation? How about for:

iii) $(x_1)^2 + (x_2)^2 = 1$

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2  
The objection to i) holds for iii) as well. –  Gerry Myerson Feb 21 '13 at 11:46
    
But is my reasoning for ii) sufficient? Is there anything more I need to say/show? –  Mathlete Feb 21 '13 at 11:47
1  
For ii), you need to know the definition of subspace, and you need to know how to apply that definition (or, better, you need to know a simple theorem about subspaces). –  Gerry Myerson Feb 21 '13 at 11:47
    
@GerryMyerson for ii) say, $x_1 = \frac{1}{3}$ and $x_2 = \frac{-1}{4}$ would I then need to show that $\frac{1}{12}$ could fit into this equation (with another value)? –  Mathlete Feb 21 '13 at 12:01
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You need to show that given any solution $(x_1,x_2)$ and any other solution $(x_3,x_4)$, any linear combination of these two solutions is also a solution. –  Gerry Myerson Feb 21 '13 at 12:03

1 Answer 1

For ii) take an arbitrary vector $(x_1,x_2)$ satisfying the equation so we can write $$(x_1,x_2)=(x_1,-\frac{3}{4}x_1)=x_1(1,-\frac{3}{4}),$$ then the set of vectors satisfying the equation is the subspace span$\{(1,-\frac{3}{4})\}$ of $\mathbb{R}^2$. Of course there is another ways to prove this.

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