To prove two vectors are linearly independent or not

Question

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Concerning the second question, I attempted to use an augmented matrix to test whether the linear system is consistent or not but failed. I also tried to turn to the original definition by writing mv+nAv=0.But I just cannot prove the coefficient must be zero. How should I complete the proof?

Answer 1

$A\begin{bmatrix}v_1 \\v_2 \\v_3\end{bmatrix}=\begin{bmatrix}cv_2-v_3 \\-cv_1+av_3 \\bv_1-av_2\end{bmatrix}$

$\exists \lambda \in \mathbb{R}^*, \exists (v_1,v_2,v_3)\neq (0,0,0)\begin{cases}cv_2-bv_3=\lambda v_1 \\ -cv_1+av_3 =\lambda v_2 \\ bv_1-av_2=\lambda v_3\end{cases} \iff \exists \lambda \in \mathbb{R}^*, \exists (v_1,v_2,v_3)\neq (0,0,0)\begin{cases}\lambda v_1 -cv_2+bv_3=0\\ -cv_1-\lambda v_2+av_3=0 \\ bv_1-av_2-\lambda v_3=0\end{cases} \iff \exists \lambda \in \mathbb{R}^*,\begin{vmatrix}\lambda & -c& b \\-c & -\lambda & a \\b & -a & -\lambda \end{vmatrix}= 0 \iff \exists \lambda \in \mathbb{R}^*,\lambda^3+(a^2+b^2+c^2)\lambda =0 \iff \exists \lambda \in \mathbb{R}^*,p(\lambda):=\lambda(\lambda^2+a^2+b^2+c^2)=0$

Now, $\forall \lambda \neq 0,p(\lambda)\neq 0$ since $(a,b,c)\neq (0,0,0)$. So, $v$ and $Av$ are linearly independent.