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preparing for an exam thats in 6 hours.

Basically, I'm stuck on trying to understand the different operations I use to finding spanning sets and linear independence.

To find a spanngin set of a set a vectors, I can reduce them and put them row echelon form, and the number of nonzero rows corresponds to the number of dimensions that it spans.

To determine linear independence, I can row reduce a matrix and if there are only nonzero rows then it is linearly independent...

These two things together being true mean we have a basis, but I dont understand how to put these together so that you have a span that isnt linearly independent, or something thats linearly independent but doesn't span....

Is my confusion making sense? I'm just typing and I don't even know what I'm trying to convey anymore...

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Linearly independent vectors that do not span:

$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$

Vectors that span, but are not linearly independent:

$\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$

I advise you to try your operations on these vectors to see what happens. It will help you understand what is going on, although 6 hours is not enough for the knowledge to sink in...

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  • $\begingroup$ To make sure...the first example doesn't span because only [1;0] and [0;1] contribute to the span, making it dimension 2, which is not 3? and could you run through why the second set spans? $\endgroup$ – Matt Mar 27 '15 at 10:16
  • $\begingroup$ @Matt the first example does not span because it can only span a 2 dimensional space (n vectors can span at most n dimensions). The second set spans because you can construct all three basic basis vectors (one in one place, zeroes in the rest) as a linear combination of the four vectors (the first two vectors are already fixed, and you can construct $[0,0,1]^T$ out of the first three. $\endgroup$ – 5xum Mar 27 '15 at 12:18

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