# Feedback on answer I wrote out for a theoretical question regarding Linear Algebra

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 $$SpB \subseteq U$$

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.

• 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. – dalta Dec 22 '18 at 17:09
• 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? – John Hughes Dec 22 '18 at 17:20
• 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. – John Hughes Dec 22 '18 at 17:21