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Under what conditions on the scalar $x$, do the following vectors form a basis for $\mathbb{C}^3$ (where $\mathbb{C}$ denotes the set of all complex numbers):- $(1,1,1), (1,x,x^2),(1,-x,x^2)$.

I have tried by forming the matrix with the given vectors and assumed that the det should be nonzero. But I'm confused because the vectors should be With complex components. Please explain your view. Thanks in advance.

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    $\begingroup$ Why exactly is confusing you about that? The scalars are complex numbers, so you can produce complex components with scalar multiplication. $\endgroup$ – amd Sep 14 '19 at 17:55
  • $\begingroup$ I don't understand your point clearly. Please explain little more. $\endgroup$ – Soumyadweep Mondal Sep 14 '19 at 18:06
  • $\begingroup$ You should explain your concerns explicitly. It appears that you’re worried about the basis vectors only having real components, but why should that be a problem? After all, the standard basis of $\mathbb C^3$ consists of $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$. Or perhaps the problem is that the equation you get has only real coefficients. Again, what’s the difficulty with that? The defining equation for $i$ also has only real coefficients. If those aren’t the issues, then explain exactly what is that’s confusing you instead of making us guess. $\endgroup$ – amd Sep 14 '19 at 18:09
  • $\begingroup$ You are right. I'm worried about the basis vectors only having real component. Why i is not involved there? And how can they generate $C^3$?? $\endgroup$ – Soumyadweep Mondal Sep 14 '19 at 18:15
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    $\begingroup$ Let's attack a slightly simpler problem first. What properties would $x$ have for $(x)^T$ to be a basis of $\mathbb{C}^1$? $\endgroup$ – Eric Towers Sep 14 '19 at 19:27
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Gaussian elimination works over every field.

Since for $x=1$ the second vector is the same as the first one, we have to exclude $x=1$. Thus we can work under the assumption that $x\ne1$. \begin{align} \begin{bmatrix} 1 & 1 & 1 \\ 1 & x & x^2 \\ 1 & -x & x^2 \end{bmatrix} &\to \begin{bmatrix} 1 & 1 & 1 \\ 0 & x-1 & x^2-1 \\ 0 & -x-1 & x^2-1 \end{bmatrix} &&\begin{aligned} R_2&\gets R_2-R_1 \\ R_3&\gets R_3-R_1\end{aligned} \\[4px] &\to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & x+1 \\ 0 & -x-1 & x^2-1 \end{bmatrix} && R_2\gets\tfrac{1}{x-1}R_2 \\[4px] &\to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & x+1 \\ 0 & 0 & 2x^2+2x \end{bmatrix} &&R_3\gets R_3+(x+1)R_2 \end{align} In order that the last row is nonzero, you need $x\ne0$ and $x\ne-1$.

Conclusion: the vectors are linearly independent for $x\ne1$, $x\ne0$, $x\ne-1$.

If you prefer, you can compute the determinant (better using the second matrix above), to get $$ (x-1)(x^2-1)-(-x-1)(x^2-1)=2x(x-1)(x+1) $$ arriving at the same conclusion.

Any value of $x$ not in the set $\{-1,0,1\}$ will yield linearly independent vectors; so $2$ is good as well as $3-2i$.

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