# Vector Spaces Linear algebra

bI've been working through some problems in my Linear Algebra course and I've come across some that have me confused. I'm not particularly good at vector spaces so some help would be greatly appreciated.

1) Suppose $V$ is a vector space, and that $U$ and $W$ are arbitrary subspaces of $V$. Determine whether the union of $U$ and $W$ is a subspace. If it is, prove it, otherwise provide a counter-example.

2) The nullspace of an $m \times n$ matrix $A$ is a subspace of $\Bbb{R}^n$. Use this statement with $m = 2$, $n = 4$, to verify that the set of vectors with coordinates $a,b,c,d$ in $\Bbb R^4$ is a subspace of $\Bbb R^4$. Determine a basis.

Needs to satisfy: $a - b + c - d = 0$; and $a + b + c + d = 0$

3) Suppose that $V$ is given as a two-dimensional vector space, which means that there is a basis of exactly two elements, say $v_1$ and $v_2$. Suppose that $v_1$ and $v_2$ are given elements in $V$ which span $V$. Show that $v_1$ and $v_2$ must be linearly independent from the definitions.

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In (2) were you given the matrix $A$? The last sentence suggests that $$A=\pmatrix{1&-1&1&-1\\1&1&1&1}\;,$$ assuming that the $-n$ was supposed to be $-b$. – Brian M. Scott Apr 11 '13 at 4:35
according to the professor, the matrix A is just a general matrix. So I am assuming that to solve these he is looking for general solutions. – Brian Apr 11 '13 at 4:40
If $A$ is a general matrix, where did the needs to satisfy conditions come from? – Brian M. Scott Apr 11 '13 at 4:40
Honestly, I have no idea. Our Professor got swapped halfway through the course and it's been a struggle to understand what is going on ever since. – Brian Apr 11 '13 at 4:47
For $(1)$ see here. – Mhenni Benghorbal Apr 11 '13 at 4:48

1. Pick any two one-dimensional subspaces of $\Bbb R^2$; is their union a subspace of $\Bbb R^2$?
3. If $v_1$ and $v_2$ are not linearly independent, there are scalars $\alpha$ and $\beta$ such that $\alpha v_1+\beta v_2=\vec 0$, but at least one of $\alpha$ and $\beta$ is non-zero. Say $\alpha\ne 0$. Then $v_1=-\frac{\beta}{\alpha}v_2$. Show that any linear combination of $v_1$ and $v_2$ is a scalar multiple of $v_2$, and conclude that $\dim V=1$, contradicting the hypothesis that $\dim V=2$.