Let \(V\) be a vector space over a subfield \(F\) of the complex numbers. Suppose \(\alpha\), \(\beta\), and \(\gamma\) are linearly independent vectors in \(V\). Prove that \((\alpha + \beta)\), \((\beta + \gamma)\), and \((\gamma + \alpha)\) are linearly independent.
Solution
Suppose \[ a(\alpha+\beta)+b(\beta+\gamma)+c(\gamma+\alpha)=0. \] Collecting coefficients of the linearly independent vectors \(\alpha,\beta,\gamma\) gives \[ a+c=0, \qquad a+b=0, \qquad b+c=0. \] From the first two equations, \(b=c=-a\). Substituting into the third gives \[ -2a=0. \] Because \(F\) is a subfield of \(\CC\), it has characteristic zero, so \(2\ne0\) and hence \(a=0\). Therefore \(b=c=0\) as well. Thus the three vectors are linearly independent.