Prove that vec a+vec b , vec b+vec c , vec c+vec a are…

Prove that $\vec a+\vec b$, $\vec b+\vec c$, $\vec c+\vec a$ are linearly independent if $\vec a,\vec b,\vec c$ are linearly independent.

Step 1: Direct proof

\[ \alpha(\vec a+\vec b)+\beta(\vec b+\vec c)+\gamma(\vec c+\vec a)=(\alpha+\gamma)\vec a+(\alpha+\beta)\vec b+(\beta+\gamma)\vec c=\vec 0. \]

By independence of $\vec a,\vec b,\vec c$: \[\alpha+\gamma=\alpha+\beta=\beta+\gamma=0\ \Longrightarrow\ \alpha+\beta+\gamma=0\ \Longrightarrow\ \alpha=\beta=\gamma=0.\qquad\blacksquare\]

Step 2: Via the triple product

Expanding by linearity gives $2^3=8$ terms. Only the two terms that use each of $\vec a,\vec b,\vec c$ exactly once survive: \[[\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]=[\vec a,\vec b,\vec c]+[\vec b,\vec c,\vec a]=2\,[\vec a,\vec b,\vec c]\neq0.\]

(Equivalently, the coefficient matrix has determinant $\det\begin{pmatrix}1&1&0\\0&1&1\\1&0&1\end{pmatrix}=2$.)

\[[\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]=2[\vec a,\vec b,\vec c]\neq0\]