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Question: If {$u,v$} is a basis for the subspace U, show that {$u+2v,-3v$} is also a basis for U

My attempt:

We must prove that {$u+2v,-3v$} spans U and is linearly independent

We know that given any $w \in U$ there exists $w = a_1(u) + a_2(v)$ where $a_1,a_2 \in \Bbb F$ and we also know that if $a_1(u) + a_2(v) = 0$ then $a_1=a_2=0$

Therefore: given any $w \in U$ there exists $w = a_1(u+2v) + a_2(-3v)$, which can be written as $w = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$ therefore {$u+2v,-3v$} spans U

By the same principle, $0 = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$,

let $a_3= 2a_1-3a_2 $, we can rewrite as $0 = (a_1)u + a_3(v)$

Therefore {$u+2v,-3v$} is also linearly independent, therefore a basis for U.

Is this correct? Thank you very much!

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  • $\begingroup$ No, neither of the arguments you present addresses the question they should. Calling $b_1=u+2v$ and $b_2=-3v$, to prove that these span $U$ you need to show that every element of $U$ (i.e., every linear combination of $u,v$) is also a linear combination of $b_1,b_2$; instead you only show that every linear combination of $b_1,b_2$ is (a linear combination of $u,v$, i.e.) an element of $U$, which is trivial. For the second part you say "Therefore" without actually having argued anything (you need to use that $u,v$ are linearly independent somewhere). The actual values of $b_1,b_2$ do matter! $\endgroup$ Jan 3 '16 at 8:09
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First, note that $\dim U=2$ so it suffices to show that $\{u+2\,v,-3\,v\}$ is linearly independent. To do so, note that $$ \lambda_1(u+2\,v)+\lambda_2(-3\,v)=0 $$ if and only if $$ \lambda_1u+(2\,\lambda_1-3\,\lambda_2)v=0\tag{1} $$ But $\{u,v\}$ is linearly independent so (1) holds if and only if \begin{array}{rcrcrcrc} 0 &=&\lambda_1\\ 0&=&2\,\lambda_1&-&3\,\lambda_2 \end{array} which holds if and only if $\lambda_1=\lambda_2=0$.

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  • $\begingroup$ Thank you for your help! So would my proof be a correct answer? $\endgroup$ Jan 3 '16 at 2:51
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    $\begingroup$ @THISISIT453 If I were grading it, I'd be inclined to say "no" - you get to the equation $0=a_1 u + a_3 v$ but you don't take the critical jump to $a_1=0$ and $a_3=0$ and solve that to see that $a_1=0$ and $a_2=0$ is the only solution. $\endgroup$ Jan 3 '16 at 3:02
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Therefore: given any $w \in U$ there exists $w = a_1(u+2v) + a_2(-3v)$, which can be written as $w = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$ therefore {$u+2v,-3v$} spans U

You should not write "given $w\in U$, there exists $w = \dots$", because $w$ exists, you are given it. I think you mean "given $w\in U$, there exists $a_1$, $a_2$ so that $w = \dots$".

Also, you are trying to show that there exist $a_1$, $a_2$ so that $w = a_1 (u+2v) + a_2(-3v)$. You do not know it yet. What you do know is that $w = b_1 u + b_2 v$. You need to play around with this so that you get $w = a_1 (u+2v) + a_2(-3v)$.

By the same principle, $0 = (a_1)u + (2a_1-3a_2)(v)$. $(2a_1-3a_2) \in \Bbb F$,

let $a_3= 2a_1-3a_2 $, we can rewrite as $0 = (a_1)u + a_3(v)$

This one is better, but still needs work. If $0 = a_1 (u +2v) + a_2 (-3v)$, then $0 = a_1 u + (2a_1-3a_2) v$. At this point, we know that since $u$ and $v$ are linearly independent, we can say $a_1 = 0$ and $a_3 := 2a_1 - 3a_2 = 0$. Remember that you wanted to show that $a_1 = a_2 = 0$ and you've only showed the first part so far. The next step is to say that because $a_3 = 0$ and $a_1=0$, we can write $a_3 = 2a_1-3a_2$ as $0 = -3a_2$, so $a_2 = 0$. Now, you know that they are linearly independent.

Hopefully seeing the work written out in the second part will give you an idea of how to work on the first part.

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There are several issues with your presentation and proof. You're talking about a subspace without mentioning the ambient space. Also, it isn't true over arbitrarily fields, e.g. it fails over a field of characteristic $3$.

I'll assume you're working over a field of characteristic $\neq 3$ and that you want to prove that $\{u,v\}$ being a basis for a vector space implies that $\{u + 2v,-3v\}$ is also a basis. Since you already know that your space is $2$-dimensional, it suffices to show that $\{u + 2v,-3v\}$ is linearly independent. So suppose $0 = a(u+2v) -3bv = au + (2a-3b)v$. Linear independence of $u,v$ implies that $a = 0$ and $2a - 3b = 0$, so $b = 0$ also, which is what we wanted to show.

A more general way to automate these kinds of arguments is to map your basis to your new suggested basis and prove that the map is an isomorphism, using your preferred method (Gauss-Jordan reduction, determinant $\neq 0$, ...).

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  • $\begingroup$ That's a very important point. I assumed OP meant $\mathbb F$ was $\mathbb R$ or $\mathbb C$, but if $\mathbb F$ has characteristic 3, $-3v = 0$. $\endgroup$
    – Snow
    Jan 3 '16 at 3:11
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To prove that $u+2v$ and $-3v$ are linearly independent we take a linear combination equated to zero, that is $$A(u+2v)+B(-3v)=0,$$ and deduce $A=B=0$.

But $A(u+2v)+B(-3v)=Au+(2A-3B)v$ then from $Au+(2A-3B)v=0$ we infer that for the coefficients $A=0$ and $2A-3B=0$ since $u,v$ are linearly independent, so $A=0$ and $B=0$.

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Use the fact that :Let $V$ is a vector space

If $\dim V=n$ then any set of $n$ vectors which are linearly independent is a basis of $V$.

So in your case it is enough to check linear independence only.

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  • $\begingroup$ Thank you for your help! So would my proof be a correct answer? $\endgroup$ Jan 3 '16 at 2:51
  • $\begingroup$ yes it would be $\endgroup$
    – Learnmore
    Jan 3 '16 at 3:01

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