Prove that vectors $\vec{x},\vec{y}$ (belonging to $\mathbb{R}^3$) are linearly dependent only if the following is true $$ \begin{vmatrix} x_1&y_1 \\ x_2&y_2 \end{vmatrix} =\begin{vmatrix} x_1&y_1 \\ x_3&y_3 \end{vmatrix}=\begin{vmatrix} x_2&y_2 \\ x_3&y_3 \end{vmatrix} = 0. $$

Could someone give any tips on how to do this?

  • $\begingroup$ I have edited your post to fix up the formatting; you can look at what I did (to see how to do it!) by clicking the "edit" button beneath your post. $\endgroup$ – Nick Peterson Jan 19 '14 at 14:10

Create a 3 by 3 matrix

$$\left[ \begin{array}{ccc} x_1 & y_1 & z_1 \\ x_2& y_2 &z_2 \\ x_3 &y_3 &z_3 \end{array} \right]$$.

Then the determinant of $z$ is given by

$$ z_3 \Bigl| \begin{array}{cc} x_1 & y_1 \\ x_2 & y_2 \end{array}\Bigr| - z_2 \Bigl| \begin{array}{cc} x_1 & y_1 \\ x_3 & y_3 \end{array}\Bigr| + z_1 \Bigl| \begin{array}{cc} x_2 & y_2 \\ x_3 & y_3 \end{array}\Bigr| $$

So given your conditions the determinant of my matrix is zero regardless of the vector $z$ so $x,y$ must be dependent. Conversely, if $x,y$ are linearly dependent then the determinant of my matrix equals zero for any vector $z$ and hence each of your determinants must be zero.



$\textbf{x}$ and $\textbf{y}$ are linearly dependent if and only if $\textbf{x}\times \textbf{y} = \textbf{0}$, and

$$ \textbf{x}\times \textbf{y} = \pmatrix{\textbf{e}_1 & \textbf{e}_2 & \textbf{e}_3\\x_1 & x_2 & x_3\\y_1 & y_2 & y_3} $$

(where $\textbf{e}_n$ is the unit vector in the $n$th dimension)

Alternate hint in absence of cross product:

What happens if $\textbf{x} = a\textbf{y}$ for some constant $a$?

  • $\begingroup$ I am not quite sure if we are allowed to use the cross product as we haven't proved it yet. $\endgroup$ – Samuli Lehtonen Jan 19 '14 at 14:02
  • $\begingroup$ See my edit, it provides an alternate hint. $\endgroup$ – Glen O Jan 19 '14 at 14:03

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