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I am new to linear algebra, and have no teacher at present. I have written out an answer to the following question, which I found very challenging, and would really appreciate any feedback. Many thanks in advance.

Let $V$ be a vector space over R. Let $v_1,v_2,v_3,u_1,u_2,u_3\in V$, so that $u_1,u_2,u_3 \in U = Sp\{v_1,v_2,v_3\}$, and so that the set $B = \{u_1-u_2,u_1+3u_3,4u_2+5u_3\}$ is linearly independent. Prove that $SpB \subseteq U$ and find the dimension of U.

My answer is the following:

It is easy to see that $B$ is a linear combination of $\{u_1,u_2,u_3\}$; in other words $B \subseteq Sp(u_1,u_2,u_3)$, and based on the information given $B \subseteq Sp(v_1,v_2,v_3)$.

$B$ is linearly independent, so $dimSpB = 3$.

Let α,β,γ be scalars in R.

We can write out B as:

$α(u_1-u_2)+β(u_1+3u_3)+γ(4u_2+5u_3)=0$

where α,β,γ are scalars in R. Because $B$ is linearly independent, α=0,β=0 and γ=0.

We can rewrite this formula as: $(α+β)u_1+(-α+4γ)u_2+(3β+5γ)u_3=0$ which shows that $u_1,u_2,u_3$ is also linearly independent. As such, $dimSp\{u_1,u_2,u_3\}=3$

So $dimSp\{u_1,u_2,u_3\}=dimSpB$. We already saw that $SpB \subseteq Sp\{u_1,u_2,u_3\}$ and therefore $SpB = Sp\{u_1,u_2,u_3\}$

$u_1,u_2,u_3 \in U = Sp\{v_1,v_2,v_3\}$ and therefore includes its own linear combinations.

Therefore $SpB \subseteq U$

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You've made a good start, but you've lost your way. Here's a line-by-line critique that I hope will help.

It is easy to see that $B$ is a linear combination of $\{u_1,u_2,u_3\}$; in other words $B \subseteq Sp(u_1,u_2,u_3)$, and based on the information given $B \subseteq Sp(v_1,v_2,v_3)$.

You have a type problem here: vectors are linear combinations of other vectors; sets of vectors are something different. What you probably meant was "It's easy to see that each element of $B$ is a linear combination of the elements of $U = \{ u_1, u_2, u_3 \}.$" But rather than asserting that it's easy to see, better would be to say "Each element of $B$ is expressed in the definition as a linear combination of the elements of $U$, hence all are in $\text{Span}(U)$.

$B$ is linearly independent, so $dimSpB = 3$. Let α,β,γ be scalars in R.

Both of these sentences are fine.

We can write out $B$ as: $α(u_1-u_2)+β(u_1+3u_3)+γ(4u_2+5u_3)=0$

This is wrong. $B$ is a set, while the left hand side of the equality above is a vector, so you're not writing out $B$.

What you can say is

"We can, by substituting the definition of each element of $B$, write out an arbitrary linear combination of elements of $B$ as $α(u_1-u_2)+β(u_1+3u_3)+γ(4u_2+5u_3)$, and we know that if this is zero, then $\alpha, \beta,$ and $\gamma$ must be zero.

That statement is at least true, although I'm not sure what you're going to do with it.

Because $B$ is linearly independent, α=0,β=0 and γ=0.

That sentence needs to be deleted.

We can rewrite this formula as: $(α+β)u_1+(-α+4γ)u_2+(3β+5γ)u_3=0$ which shows that $u_1,u_2,u_3$ is also linearly independent.

What you can really say is that if $(α+β)u_1+(-α+4γ)u_2+(3β+5γ)u_3=0$, then $\alpha, \beta, \gamma$ must all be zero. To show that the elements of $U$ are independent, you need to show that if $au_1 + bu_2 + cu_3 = 0$, then $a, b, c$ are all zero. You haven't done that at all.

As such, $dimSp\{u_1,u_2,u_3\}=3$

If you had been able to conclude what you wrote in the previous sentence, this one would be OK.

So $dimSp\{u_1,u_2,u_3\}=dimSpB$.

Same for this one.

We already saw that $SpB \subseteq Sp\{u_1,u_2,u_3\}$ and therefore $SpB = Sp\{u_1,u_2,u_3\}$

And this.

$u_1,u_2,u_3 \in U = Sp\{v_1,v_2,v_3\}$ and therefore includes its own linear combinations.

I have no idea what that sentence means, because the "its" refers to something that I'd expect to be the subject of the sentence...but the subject is $u_1, u_2,_3$, which is plural.

Therefore $SpB \subseteq U$

Back when you said "We can wrote out $B$ as ...", you could have re=gathered terms to say that each element of $\text{span}(B)$ has the form $(\alpha + \beta) u_1 + (-\alpha + 4\gamma)u_2 + ...$, which is evidently a linear combination of elements of $U$, hence in $\text{Span}(U)$, and you'd have been done.

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  • $\begingroup$ Thank you: this is invaluable help for me! One remaining question: if my reasoning was incorrect, then how do I calculate the dimension of U based on the information provided? It is part of the question, so there must be a way. Yet, without a way of checking u1,u2,u3 for linear independence, I am lost as to how to compute it. $\endgroup$ – dalta Dec 22 '18 at 17:09
  • $\begingroup$ I already told you: "To show that the elements of $U$ are independent, you need to show that if $au_1+bu_2+cu_3=0$, then $a,b,c$ are all zero." So assume that $au_1+bu_2+cu_3=0$, and see whether you can turn this into a statement about linear combinations of the $b$s, which you *know* are independent. Hint: Can you write $u_1$ as a linear combination of the $b$s? $\endgroup$ – John Hughes Dec 22 '18 at 17:20
  • $\begingroup$ There's also a far more direct argument involving dimension, but may use ideas you have not yet encountered, so I'm avoiding that for now. $\endgroup$ – John Hughes Dec 22 '18 at 17:21

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