1.6 Exercise 11
Let \(u\) and \(v\) be distinct vectors of a vector space \(V\). Show that if \(\{u, v\}\) is basis for \(V\), and \(a\) and \(b\) are non-zero scalars, then both \(\{u+v, au\}\) and \(\{au, bv\}\) are also bases for \(V\).


Proof:
To prove that \(\{u+v, au\}\) is basis, we need to show that the set of vectors are linearly independent and that \(span(\{u+v, au\}) = V\). Let \(a_1, a_2\) be any two scalars. Then we want to prove that the following equation

$$ \begin{align*} a_1(u+v) + a_2(au) = \bar{0} \end{align*} $$

is only satisfied by the trivial solution. We’ll re-arrange the terms as follows,

$$ \begin{align*} \bar{0} &= a_1u + a_1v + aa_2u \\ &= (a_1 + aa_2)u + a_1v \\ &= (a_1 + aa_2)u + a_1v. \end{align*} $$

From this, we can see that the only solution is the trivial solution since \(\{u,v\}\) are linearly independent. Moreover, since \(\{u,v\}\) is a basis, then its dimension is 2. Therefore, \(\{u+v, au\}\) is a basis.

Similarly, to see that \(\{au, bv\}\), we’ll prove the following equation’s solution is the trivial solution only by re-arranging the terms,

$$ \begin{align*} \bar{0} &= a_1(au) + a_2(bv) \\ \bar{0} &= (a_1a)u + (a_2b)v. \end{align*} $$

This is again a linear combination of two linearly independent vectors and so the trivial solution is the only possible solution to this equation. Since the set has 2 vectors, then it’s a basis as well. \(\blacksquare\)

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