I am stuck on the following problem:

Prove that the vectors $\vec{\text{u}}_1$, $\vec{\text{u}}_2$, and $\vec{\text{u}}_3$ are linearly independent if and only if $$ \det \begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \neq 0 $$

I'm having trouble choosing which direction to go about. I'm aware of the following:

  • The above determinant is equivalent to the triple product of said vectors: $$ \vec{\text{u}}_1 \cdot (\vec{\text{u}}_2 \times \vec{\text{u}}_3) $$
  • The three above vectors are linearly independent if the equation $$ a_1 \vec{\text{u}}_1 + a_2 \vec{\text{u}}_2 + a_3 \vec{\text{u}}_3 = 0 $$ has only the trivial solution.

  • Any two vectors are linearly independent if their dot product is equal to 0.

My first attempt was to use the three combinations of dot products to create the following system : \begin{gather*} a_1 x_1 x_2 + a_1 y_1 y_2 + a_1 z_1 z_2 = 0 \\ a_2 x_1 x_3 + a_2 y_1 y_3 + a_2 z_1 z_3 = 0 \\ a_3 x_2 x_3 + a_3 y_2 y_3 + a_3 z_2 z_3 = 0 \end{gather*} and somehow work it into the determinant, but I feel like this isn't the correct way to do it.

Another idea was to use Sarrus' rule on the determinant to obtain a new equation through which I could work my way up to the second bullet point, but that doesn't seem to make any sense.

Maybe I can't see the forest for the trees; I'm not really good at linear algebra proofs. I'm just looking for a hint to go in the right direction.

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    $\begingroup$ Don't know that you are allowed to use that, but it's well known that the corresponding Homogeneous linear system with non-0 determinant has only the trivial zero solution. $\endgroup$ – dxiv Sep 25 '16 at 21:47
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    $\begingroup$ Intuitively, $ \vec{\text{u}}_1 \cdot (\vec{\text{u}}_2 \times \vec{\text{u}}_3) \neq 0$ means that $\vec{\text{u}}_1$ is not orthogonal to $\vec{\text{u}}_2 \times \vec{\text{u}}_3$. In other words, $\vec{\text{u}}_1$ is not in the same plane defined by $(\vec{\text{u}}_2, \vec{\text{u}}_3)$, which implies that $\vec{\text{u}}_1$ cannot be expressed as a linear combination of the other two vectors. $\endgroup$ – trang1618 Sep 25 '16 at 22:00

You should write the vectors as column vectors. A linear relation between $\vec u_1, \vec u_2$ and $\vec u_3$ can be written matricially as $$A\begin{bmatrix} a_1\\a_2\\a_2\end{bmatrix}=\begin{bmatrix} x_1 & x_2&x_3\\ y_1 & y_2&y_3\\z_1 & z_2&z_3\end{bmatrix}\begin{bmatrix} a_1\\a_2\\a_3\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$$ Now multiply both sides on the left by the adjugate matrix of $A$: $$\operatorname{adj}(A)\,A\begin{bmatrix} a_1\\a_2\\a_2\end{bmatrix}= \det A\: I_3\begin{bmatrix} a_1\\a_2\\a_2\end{bmatrix}=\det A\begin{bmatrix} a_1\\a_2\\a_2\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$$ As $\det A\neq 0$, this implies $$\begin{bmatrix} a_1\\a_2\\a_2\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}.$$

  • $\begingroup$ Is $u_1=(x_1,y_1,z_1)$? $\endgroup$ – snulty Sep 26 '16 at 0:26
  • $\begingroup$ Absolutely. $ $ $\endgroup$ – Bernard Sep 26 '16 at 1:03
  • $\begingroup$ Ha, that's how I tried next, except I had forgotten about transposing the coordinate matrix. Thanks! $\endgroup$ – AgentRev Sep 26 '16 at 3:38

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