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How do I show that $S=\{u_1,u_2\}$ is a basis for the solution space $Ax=0$ without doing Elementary.Row.Operations?

Given the solution space of $Ax=0$ has dimension 2.

$$A=\begin{pmatrix} 3 & -1 & -2 & -4\\ 0 & 1 & -1 & 1\\ -1 & 1 & 0 & 2\\ -2 & 1 & 1 & 3 \end{pmatrix}$$

$$u_1=\begin{pmatrix} 1\\ 1\\ 1\\ 0 \end{pmatrix} ,u_2=\begin{pmatrix} 1\\ -1\\ 0\\ 1 \end{pmatrix}$$

Ok, so my guess is that, since $u_1$ and $u_2$ are linearly independent, and there are two vectors in the basis, that fulfils one of the conditions for S to be a basis for V.

Now I need to prove that S spans V, but how do I do that without doing any E.R.Os?

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2 Answers

up vote 1 down vote accepted

This is a standard (and fundamental) result which may be of interest to you.

Theorem: Let $V$ be an $n$-dimensional vector space. Let $\mathcal{B}$ be a set of $n$ vectors in $V$. Then the following are equivalent

  1. $\mathcal{B}$ is a basis for $V$.
  2. $\mathcal{B}$ is linearly independent.
  3. $\mathcal{B}$ is a spanning set of $V$.

You have a set of two linearly independent in a subspace of dimension $2$. The vectors are thus necessarily a basis.

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Do I have to show $u_1,u_2$ form a solution for $Ax=0$? –  Yellow Skies Nov 4 '12 at 11:58
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You need to show that they actually lie in the kernel, yes. If you can show that they're linearly independent and that they're both in the kernel then you're done. –  EuYu Nov 4 '12 at 11:58
    
I see~ Thanks you two –  Yellow Skies Nov 4 '12 at 12:00
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Considering that your problem is only to show that $S$ is a basis, if you know that the dimension of the kernel space is 2 and if you have 2 linearly independent vectors, so they will span the kernel. By the other hand, if you really want to show this statement, you have to take an arbitrary element of the kernel and write it as linear combination of vectors in $S$.

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Oh I SEE! So I just need to show that S is linearly independent, and S has the same dimension space, and I'm done? –  Yellow Skies Nov 4 '12 at 11:52
    
I'm supposing that you already know the dimension of kernel space. Now, read the answer given by EuYu. –  Sigur Nov 4 '12 at 11:57
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