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Consider the complex vector space $\mathbb{C}^3$ and the subset $$S = \{(1, 0, i), (i, 2, -3), (2-i, 1+i, i)\}$$

Let $\textsf V$ denote $\operatorname{span}(S)$. Find a subset of $S$ that forms a basis for $\textsf V$.

How do I find the subset for something? I can put the vectors into matrix form and make them equal to $x, y, z$ respectfully but from there, I don't know how to proceed.

Am I missing a definition and and a series of steps? I know that a basis in simple terms is :

The minimal amount of vectors that are linearly independent and spans $\textsf V$.

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    $\begingroup$ $S$ should be a subset of $V$ consisting of $3$ vectors. Do you know how to tell if the $3$ are linearly independent? $\endgroup$ – J. W. Tanner Sep 12 at 1:27
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How do you know the linear dependence between those vectors? Well, I'll use the definition, suppose there exists complex scalars $a_1,a_2,a_3$ such that $$\begin{pmatrix} 0\\0\\0 \end{pmatrix} = a_1\begin{pmatrix} 1\\0\\i \end{pmatrix}+a_2\begin{pmatrix} 2-i\\1+i\\i \end{pmatrix}+a_3\begin{pmatrix} i\\2\\-3 \end{pmatrix} = \begin{pmatrix} 1&2-i&i\\0&1+i&2\\i&i&-3 \end{pmatrix}\begin{pmatrix} a_1\\a_2\\a_3 \end{pmatrix} \tag{1}$$ then, finding the row reduced echelon form of this matrix, yields that $$\begin{pmatrix} 1&0&-1+4i\\0&1&1-i\\0&0&0 \end{pmatrix}\begin{pmatrix} a_1\\a_2\\a_3 \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}$$ Then, the system has infinite solutions, so $ S $ cannot be linearly independent. We quickly see that $$\begin{pmatrix} a_1\\a_2\\a_3 \end{pmatrix} = \begin{pmatrix} 1-4i\\-1+i\\1 \end{pmatrix}$$ is a solution of that system. But also, is a solution of the original system, $(1)$, that is : $$\begin{pmatrix} 0\\0\\0 \end{pmatrix} = (1-4i)\begin{pmatrix} 1\\0\\i \end{pmatrix}+(-1+i)\begin{pmatrix} 2-i\\1+i\\i \end{pmatrix}+1\begin{pmatrix} i\\2\\-3 \end{pmatrix}$$ or $$(i,2,-3)=(-1+4i)(1,0,i)+(1-i)(2-i,1+i,i)$$ Thus, a basis for $\textsf V$ consist of $$S\setminus \{ (i,2,-3) \} = \{ (1,0,i),(2-i,1+i,i) \}$$

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  • $\begingroup$ Does it need to be mentioned that the two remaining vectors are linearly independent? $\endgroup$ – J. W. Tanner Sep 12 at 2:12
  • $\begingroup$ It is a question? $\endgroup$ – Azif00 Sep 12 at 2:12
  • $\begingroup$ Yes; to be pedantic it should be mentioned, but maybe it is so apparent that it goes without saying $\endgroup$ – J. W. Tanner Sep 12 at 2:13
  • $\begingroup$ It is clear, since one is not a scalar multiple of the other. That's right. $\endgroup$ – Azif00 Sep 12 at 2:14
  • $\begingroup$ I agree; it was merely a comment $\endgroup$ – J. W. Tanner Sep 12 at 2:14
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The span of $S$ is $2$-dimensional because $(i,2,-3)-(1-i)(2-i,1+i,i)-(4i-1)(1,0,i)=0.$

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    $\begingroup$ Another way to show that the vectors in $S$ are not linearly independent would be to show that the determinant of the "vectors in matrix form" is $0$ $\endgroup$ – J. W. Tanner Sep 12 at 1:59

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