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I'm having trouble writing this a proof for linear algebra.

Let $u$ and $v$ be vectors in $\mathbb R^3$. Let $V=\{au+bv \mid a,b \in \mathbb R\}$. Prove that if $x$ and $y$ are in $V$, then $x+y$ is in $V$.

Proof: Let $x,y \in V$. Then $x,y \in au+bv$ and $x,y \in \mathbb R$. We have $$au+bv=xu+yv =(x+y)(u+v)$$

I know I went wrong somewhere but I don't know how to fix it or how I should approach it. I feel like it should be very simple but I'm a little rusty on my proof writing.

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  • $\begingroup$ I fixed the $\LaTeX$ in your question. For more on how to use math markup on this site, see this page. $\endgroup$ – André 3000 Aug 24 '17 at 20:49
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If $\mathbf{u}$ and $\mathbf{v}$ are $vectors$ in $\mathbb{R}^{3}$, then the set $V = \{a\mathbf{u}+b\mathbf{v} | a, b \in \mathbb{R}\}$ is the set of their linear combinations (vectors you get by addition and/or scaling of $\mathbf{u}$ and $\mathbf{v}$), which means that, if $\mathbf{x}$ and $\mathbf{y}$ are in $V$, they must also be vectors. We know from the statement that if $\mathbf{x}, \mathbf{y} \in V$, then each of them is a linear combination of vectors $\mathbf{u}$ and $\mathbf{v}$: $$\mathbf{x} = m \mathbf{u} + n \mathbf{v}$$ and $$\mathbf{y} = p \mathbf{u} + q \mathbf{v},$$ where $m,n,p$ and $q$ are real numbers. If we substitute this in $\mathbf{x}+\mathbf{y}$, we get $$\mathbf{x}+\mathbf{y} = m \mathbf{u} + n \mathbf{v} + p \mathbf{u} + q \mathbf{v} = (m+p)\mathbf{u} + (n+q)\mathbf{v}.$$ Since $m,n,p$ and $q$ are real numbers, the sums $m+p$ and $n+q$ are also real, which means that $(m+p)\mathbf{u} + (n+q)\mathbf{v} \in V$, that is, $\mathbf{x}+\mathbf{y} \in V$.

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Your equalities are wrong and has no sense. Let $x,y\in V$. Then, there is $a,b,c,d\in \mathbb R$ s.t. $$x=au+bv\quad \text{and}\quad y=cu+dv.$$

Then $$x+y=au+bv+cu+dv=\underbrace{(a+c)}_{\in \mathbb R}u+\underbrace{(b+d)}_{\in\mathbb R}v\in V,$$ what prove the claim.

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