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I found the following question: Let V be a space of all vectors in $R^2$ that satisfy the following conditions:

a) $x_1 + x_2 + x_3 + x_4 + x_5 = 0$
b) $x_1 - x_2 + 2x_3 -2x_4 + 3x_5 = 0$

Find a basis for V.

I do know how to find a basis that satisfies only 1 condition, for example condition a: that would consist of $ (1, 1, 1, 1, -4)^T , (1, 1, 1, -4, 1)^T, (1, 1, -4, 1, 1)^T , (1, -4, 1, 1, 1)^T $, but I can't get this to work with 2 conditions. I realize this question is the same as finding the nullspace of the 2x5 matrix

$\begin{bmatrix}1 & 1 & 1 & 1 & 1\\1 & -1 & 2 & -2 & 3\end{bmatrix}$

but I didn't succeed in getting an answer to the question that way. I hope someone could help me out with this.

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    Have you tried row-reducing your matrix? What *have* you tried? – 2017-01-30
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    Yes I've tried that. Row1-->Row1 - 2Row2 gives $x_2 - x_3 + 3x_4 -2x_5 = 0$, and then I tried adding 2* row 1 of row 2, which gives $x_1 + 3x_3 + 4x_4 - x_5 = 0$, but that doesn't help me either I think. I also wrote $x_3$ and $x_1$ in terms of $x_2 , x_4 and x_5$ and then added $x_1$ and $x_3$ which gave me two new equations: $x_1 + x_3 = -5x_2 - 7x_4 + 3x_5$ – 2017-01-30
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    $x_1 + x_3 = -x_2 - x_4 - x_5$ so $-4x_2 - 6x_4 + 4x_5 = 0$ And that's about all I tried – 2017-01-30
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    You are on the right track. Think carefully about what a basis is in this case, and what the vectors satisfying it have to look like. – 2017-01-30
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    Well, they should be linearly independant. I've thought about it for a long time before posting this, but I've no idea what to do from here. – 2017-01-30

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