Problem. Let \(V\) be a vector space over \(\mathbb{R}\) and let \((v_{1},v_{2},v_{3})\) be a linearly independent sequence of vectors from \(V\). Determine for which values \(a\in\mathbb{R}\) the sequence \[S=(v_{1}+av_{2},\;v_{2}+av_{3},\;v_{3}+av_{1})\] is also linearly independent. Prove your answer.
Proof. Let us recall the theorem which states that a sequence of vectors is linearly independent if and only if the only way to reach \(0\) as a linear combination of it is by taking all of the scalars as zero.
Let us examine the following equation
\[\alpha_{1}\left(v_{1}+av_{2}\right)+\alpha_{2}\left(v_{2}+av_{3}\right)+\alpha_{3}\left(v_{3}+av_{1}\right)=0\] \[\alpha_{1}v_{1}+\alpha_{1}av_{2}+\alpha_{2}v_{2}+\alpha_{2}av_{3}+\alpha_{3}v_{3}+\alpha_{3}av_{1}=0\] \[\left(\alpha_{1}+\alpha_{3}a\right)v_{1}+\left(\alpha_{1}a+\alpha_{2}\right)v_{2}+\left(\alpha_{2}a+\alpha_{3}\right)v_{3}=0\]
And from the same theorem, since \((v_{1},v_{2},v_{3})\) is linearly independent, we know that \[\begin{cases} \left(\alpha_{1}+\alpha_{3}a\right)=0\\ \left(\alpha_{1}a+\alpha_{2}\right)=0\\ \left(\alpha_{2}a+\alpha_{3}\right)=0 \end{cases}\]
We will use this linear system to determine which \(a\) values give us \(\alpha_{1}=\alpha_{2}=\alpha_{3}=0\), i.e. that the only solution is the trivial solution.
\[\begin{pmatrix}\begin{array}{ccc|c} 1 & 0 & a & 0\\ a & 1 & 0 & 0\\ 0 & a & 1 & 0 \end{array}\end{pmatrix}\overset{R_{2}\rightarrow R_{2}-aR_{1}}{\Longrightarrow}\begin{pmatrix}\begin{array}{ccc|c} 1 & 0 & a & 0\\ 0 & 1 & -a^{2} & 0\\ 0 & a & 1 & 0 \end{array}\end{pmatrix}\overset{R_{3}\rightarrow R_{3}-aR_{2}}{\Longrightarrow}\begin{pmatrix}\begin{array}{ccc|c} 1 & 0 & a & 0\\ 0 & 1 & -a^{2} & 0\\ 0 & 0 & 1+a^{3} & 0 \end{array}\end{pmatrix}\]
\(\left(1+a^{3}\right)\alpha_{3}=0\) means \(\alpha_{3}=0\) or \(a^{3}=-1\iff a=-1\).
Considering this, \(\alpha_{2}-a^{2}\alpha_{3}=0\) means that if \(\alpha_{3}=0\) then \(\alpha_{2}=0\) and if \(a=-1\) then \(\alpha_{2}=\alpha_{3}\).
Similarly, \(\alpha_{1}+a\alpha_{3}=0\) means that if \(\alpha_{3}=0\) then \(\alpha_{1}=0\) and if \(a=-1\) then \(\alpha_{1}=\alpha_{3}\).
Which means that if \(a\neq-1\) the set of solutions is \(\left\{ \begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}\right\}\) and if \(a=-1\) it’s \(\left\{ \begin{pmatrix}x\\ x\\ x \end{pmatrix}\mid x\in\mathbb{R}\right\}\), which means that the sequence is linearly independent for \(a\in\mathbb{R}\setminus\{-1\}\). ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.