Change of basis independency of vectors

Question

Let $V$ be a vectorspace over $\mathbb{R}$ and let $\mathcal{B}:=(v_1,v_2)$ be a basis of $V$.

a) Let $v_1',v_2'\in V$ given by $_{\mathcal{B}}[v_1']=\begin{pmatrix}1\\1\end{pmatrix}$ and $_{\mathcal{B}}[v_2']=\begin{pmatrix}1\\-1\end{pmatrix}$.

Show that $\mathcal{B}':=(v_1',v_2')$ is a basis of $V$ too.

Let $v_1=\begin{pmatrix}a\\b\end{pmatrix}$ and $v_2=\begin{pmatrix}c\\d\end{pmatrix}$, then $v_1'=\begin{pmatrix}a+c\\b+d\end{pmatrix}$ and $v_2'=\begin{pmatrix}a-c\\b-d\end{pmatrix}$.

I tried to use the determinant argument, but then I run into a lot of

'if a=0, then b$\not=$0'-shenanigans.

Is there a better way that actually uses the (for me) new notation '$_{\mathcal{B}}[v_1']$'?

Since it is quite obvious that (1,1) and (1,-1) are independant, but does this mean that then $v_1',v_2'$ are too?

Answer 1

The notation $_{\mathcal{B}}[v_1']=\begin{pmatrix}1\\1\end{pmatrix}$ means that the coordinates of $v_1'$ in the basis $\mathcal{B}=(v_1,v_2)$ are $\begin{pmatrix}1\\1\end{pmatrix}$, or equivalently, that $v_1'=v_1+v_2$. Similarly, $_{\mathcal{B}}[v_2']=\begin{pmatrix}1\\-1\end{pmatrix}$ means that $v_2'=v_1-v_2$.

Using this, it becomes easy to see that $v_1',v_2'$ span $V$; indeed, notice that $v_1=\frac{v_1'+v_2'}{2}$ and $v_2=\frac{v_1'-v_2'}{2}$, so that $V=\operatorname{span}(v_1,v_2)\subset \operatorname{span}(v_1',v_2')$.

You can also show that they must be linearly independent, since $$0=\alpha_1v_1'+\alpha_2v_2'=(\alpha_1+\alpha_2)v_1+(\alpha_1-\alpha_2)v_2$$ implies that $\alpha_1+\alpha_2=0=\alpha_1-\alpha_2$, and thus that $\alpha_1=0=\alpha_2$.

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Another way to see the result is to observe that the function $$\varphi : v\in V\mapsto \ _\mathcal{B}[v]\in \Bbb R^2$$ is an isomorphism of vector spaces. This implies that $(v_1',v_2')$ is a basis of $V$ if and only if $( _\mathcal{B}[v_1'], \ _\mathcal{B}[v_2'])$ is a basis of $\Bbb R^2$. This should answer the question in your last sentence.