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Suppose that $\{u,v,w\}$ is a linearly independent set of vectors in R50.

(a) Show that $u + 2v + 3w$ is in $Span\{u,v,w\}$.

(b) Show that $u + 2v + 3w$ is in $Span\{u−v,u−2v + w,v + w\}$.

I struggle with vector spaces because they are so abstract, can anyone help me with these?

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    What have you tried and where did you get stuck? Do you know the *definition* of Span? – 2017-02-27
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    The span is the set of all linear combinations of the basis vectors. $u+2v + 3w$ is a liner combination of $u,v,w$ – 2017-02-27
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    So, is there a way to prove that u+2v+3w is a linear combination of u,v,w, or can I just say that this is trivial? – 2017-02-27
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    @AmaC It depends on the definition you use in your class. I've seen span$\{u,v,w\}$ defined as the smallest subspace containing $u$, $v$, and $w$. If you use that definition you will have to say more than 1 sentence. But if you use the definition that span$\{u,v,w\}$ is the set of all linear combinations, then yeah, you can just say "$u+2v+3w$ is a linear combination of $\{u,v,w\}$ and thus in span$\{u,v,w\}$". – 2017-02-27

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I'll help you do (a) and I'll give you a hint for (b).

The very definition of span$\{u,v,w\}$ means it's the set of elements that look like $\{c_{1} u + c_{2} v + c_{3} w \}$ where the $c_{i}$ are real numbers. For example, $1u + 3v - 17w$ is in the set span$\{u,v,w\}$. Also, $-\sqrt{2}u - \pi v + 0.656 w$ is in span$\{u,v,w\}$. Basically, span$\{u,v,w\}$ is the set of every linear combination of the vectors $u, v,$ and $w$. Note that $2u + 5w$ is also in span$\{u,v,w\}$ since it can be written as $2u + 0v + 5w$. So a vector $s$ is in span$\{u,v,w\}$ if we can write $s$ as $c_{1}u + c_{2}v + c_{3}w$ for some real numbers $c_{1}, c_{2},$ and $c_{3}$. So, to answer (a), clearly by the above explanation, $u + 2v + 3w$ is in the set span$\{u,v,w\}$ since it's a linear combination of $u, v,$ and $w$.

For (b), here is the hint: By the definition of span as above, you have to find real numbers $c_{1}$, $c_{2}$, and $c_{3}$ so that $u + 2v + w = c_{1} (u-v) + c_{2}(u - 2v + w) +c_{3}(v+w) $. Extra hint: distribute the $c_{i}$ on the right hand side of the equation and then factor it out to look like $(number1)u + (number2)v + number(3)w$. Then since we have $u + 2v + w = (number1)u + (number2)v + number(3)w$, that means $1 = number1$, $2 = number2$, and $1 = number3$ (i.e., the coefficients of $u, v,$ and $w$ are equal). Use these three equations to solve for $c_{1}$, $c_{2}$, and $c_{3}$.

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    This makes a lot of sense, thank you very much! – 2017-02-27
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    @AmaC you are very welcome! :) – 2017-02-27