How to determine the basis with given vectors? So I tried to solve this problem, but I can't seem to find an answer.
So the question is: "Using determinants, indicate the sets of values of $x$, $y$, and $z$ for which the vector sequence is a basis.
$$
S = ((0, z, -y), (-z, 0, x), (y, -x, 0))
$$
I tried to put these three vectors in columns to calculate the determinant and it gives me $0$. But I did something wrong because the determinant can't be $0$, otherwise, this couldn't be a basis. Can someone help me, please?
 A: Have you considered the possibility that you have not made a mistake?
It seems like you already know that if the determinant of the matrix $\begin{pmatrix} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{pmatrix}$ is $0,$ then the column vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are not linearly independent, hence cannot be the elements of a basis.
You have also calculated this determinant, and found that it was $0$. So, you should conclude that for any $x, y, z$, the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are always linearly dependent, hence they never form a basis.
Indeed, we can show this directly. First, suppose that $z = 0$. Then, $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -y \end{pmatrix}$ and $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ x \end{pmatrix}$ are clearly linearly dependent, hence the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ do not form a basis.
Now, suppose $z \neq 0.$ Then, $-xz^{-1}$ and $-yz^{-1}$ are well-defined scalars. Note that we have
$$ (-xz^{-1})\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix} + (-yz^{-1})\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix} = \begin{pmatrix} 0 \\ -x \\ xyz^{-1} \end{pmatrix} + \begin{pmatrix} y \\ 0 \\ -xyz^{-1} \end{pmatrix} = \begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}.$$ So, the vectors $\begin{pmatrix} 0 \\ z \\ -y \end{pmatrix}$, $\begin{pmatrix} -z \\ 0 \\ x \end{pmatrix},$ and $\begin{pmatrix} y \\ -x \\ 0 \end{pmatrix}$ are linearly dependent, hence do not form a basis.
Having considered both possibilities for $z$, we've shown that regardless of what $x, y, z$ are, the three vectors do not form a basis. So, your calculation was correct.
