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What is the span of $(1, 1, 1), (0, -1, 1), (0, 0, -1) \in \mathbb R^3$?

Supposing we haven't covered linear in/dependence, can we solve the problem as done below?

The span is a set of all systems:

$$ \left\{ \begin{array}{c} a+b\cdot 0 + c\cdot 0 \\ a-b + c \cdot 0 \\ a+b -c \end{array} \right. $$

where $a, b, c \in \mathbb R$.

Suppose $(x, y, z) \in \mathbb R^3$ and

$$ \left\{ \begin{array}{c} a+b\cdot 0 + c\cdot 0 = x \\ a-b + c \cdot 0 = y\\ a+b -c = z \end{array} \right. $$

Then $b = x – y$ and $c = x + x – y – z = 2x -y – z$.

So,

$x(1, 1, 1) + (x – y)(0, -1, 1) + (2x -y – z)(0, 0, -1)$

$= (x, x, x) + (0, y – x, x – y) + (0, 0, -2x + y + z)$

$= (x, y, z)$

Thus the given set of vectors spans $\mathbb R^3$.

Em.
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1 Answers1

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Yes. Indeed, using the first equation constrains the first coordinate, it is clear you can use the second equation to come up with any pair of first coordinates, and the third - to come up with any triple, which you did.

gt6989b
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