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Let $ S = \{ (1,1,0), (0,1,1), (1,0,1) \} \subset \Bbb R^3 .$

a) Show that S spans $\Bbb R^3$

b) Show that S is a basis for $\Bbb R^3 $

I cannot use the rank-dimension method for (a). Is it possible to show via combination of 3 vectors? What would the final equations be like? I tried using this method but I don't think it's the final answer.

$x+z = a \\ x+y=b\\ y+z=c$

So what do I do from here.

I'm not sure how to do (b).

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up vote 3 down vote accepted

Let $$v\in \Bbb R^3$$ Then v = (v1,v2,v3). Let s1 = (1,1,0), s2 = (0,1,1), s3 = (1,0,1).

So by row reducing the matrix (s1 s2 s3 v), we get that $$v = ((v_1+v_2-v_3)/2)*s_1 + ((-v_1+v_2+v_3)/2)*s2 + ((v_1-v_2+v_3)/2)*s_3$$

And since v was chosen arbitrarily this means that $$S\ \text{spans}\ \Bbb R^3$$

Now all you need to do is show that S is linearly independent. (Hint: show that there is only one solution to this (s1 s2 s3 x) where x is the zero vector.

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to show $S$ spans $\mathbb{R}^3$, consider the set of all linear combinations of $S$ - make a matrix with $S$ as column vectors, and all linear combinations of $S$ are the image of $\mathbb{R}^3$ under left multiplication by $A$. So claim that $S$ spans $\mathbb{R}^3$ implies that for each $(a,b,c)^T \in \mathbb{R}^3$, there is a solution to the system $S (x,y,z)^T = (a,b,c)^T$. Solve the system directly, there exists a unique solution.

The uniqueness implies it's a basis also (since uniqueness implies linear independence of the elements of $S$).

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