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Let $V$ be a subspace of $R^n$ of dimension $3$. If $β = \{v_1,v_2,v_3\}$ is a basis of $V$ , then is $γ = \{cv_1,v_1 +v_2,v_1+v_2+v_3\}$ necessarily a basis of $V$ for $c \neq 0$.

I think because the vectors in $γ$ are linearly independent and thus should be a basis for a subspace of dimension 3. Is this thought right?

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    $\begingroup$ Thats correct, but you need to show that every element of $V = \text{span} \beta$ are linear combination of the new basis. However, please use MathJax $\endgroup$ – Sou Nov 17 '17 at 0:14
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First we show that $\text{span}(\gamma) \subseteq \text{span}(\beta) = V$. Let $\vec{v} \in \text{span}(\gamma)$.

We have that $\vec{v} = r_1(cv_1)+r_2(v_1+v_2)+r_3(v_1+v_2+v_3)$ for some $r_1,r_2,r_3 \in \Bbb{R}$ by definition of a linear combination. Then rewriting this, we get $$\vec{v} = (cr_1+r_2+r_3)v_1+(r_2+r_3)v_2+r_3v_3$$ and we're done with that relation as this is a linear combination of $\beta$.

To show that $\text{span}(\beta) \subseteq \text{span}(\gamma)$, we take an element $\vec{w} \in \text{span}(\beta)$ and show it can be written as a linear combination of the elements in $\gamma$.

Similarly, we know $\vec{w} = s_1v_1+s_2v_2+s_3v_3$ for some $s_1,s_2,s_3\in\Bbb{R}$. We then note that $$s_1v_1+s_2v_2+s_3v_3=(\frac{s_1}{c}-\frac{s_2}{c})(cv_1)+(s_2-s_3)(v_1+v_2)+s_3(v_1+v_2+v_3)$$ (which can simply be checked by algebra) and this is a linear combination of the elements of $\gamma$.

Thus we have shown that $\text{span}(\gamma) \subseteq \text{span}(\beta)$ and $\text{span}(\beta) \subseteq \text{span}(\gamma)$, so we can then conclude that $\text{span}(\gamma) = V$.

Finally, since $\text{dim}( V ) = 3$, we know that $\gamma$ must be a basis since it both spans $V$ and contains 3 elements. This is due to the theorem which states that given a subset $S\subseteq W$ ($W$ a vector space) with $|S|=n$ and with $\text{dim}(W)=n$, we have that $S$ is linearly independent iff it spans $V$.

EDIT:

Conversely, you can prove that $\gamma$ is linearly independent and use the same theorem.

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