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?

  • 1
    $\begingroup$ 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

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 $V$.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.