# to check linear span of complex number

consider the set $\{ (1,0,-i),(1+i,1-i,1),(i,i,i)\}$ of three vectors from $\mathbb{C}^3$. which of the following is true?

a) linear span of set is of dimension $1$
b) linear span of set is of dimension $2$
c) each element of $\mathbb{C}^3$ can be generated as real combination of elements of set
d) set is basis of $\mathbb{C}^3$

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You can row reduce as you are probably accustomed to. The vector $(i,i,i)$ is kind of ugly, I would replace it by $(1,1,1)$. That's OK, $(1,1,1)$ is $(i,i,i)$ multiplied by a non-zero constant. –  André Nicolas Feb 3 '13 at 7:09

If you set a matrix whose rows are the vectors noted above and find the Reduced Row Echelon Form of the matrix you will see that the vectors establish a basis for $\mathbb C^3$. In fact: $$\begin{pmatrix} 1 & 0 & -i\\ 1+i & 1-i & 1\\ i&i &i \\ \end{pmatrix}\approx\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$$

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I like this answer! +1 –  amWhy Feb 3 '13 at 13:21

$a. (1,0,-i) + b.(1+i,1-i,1) +c.(i,i,i)= (0,0,0)$

$a +b(1+i) +ci=0\tag{1}$

$b(1-i) + ci =0 \tag{2}$

$-ai +b + ci=0\tag{3}$ $(2) -(3) \implies a=b$ from $(1) a=c =0$ thus $a=b-c=0$

thus set is a basis

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