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Let $\vec{v} ∈ S$ then $\vec{v} = (-z_2,z_2,z_3)$, by definition of $S$, $z_1=-z_2$.

This means that I can write to any vector of S as a linear combination of $(-1,1,0), (0,0,1),(-i,i,0)$ and $(0,0,i)$ so this vectors spanning $S$. Is this correctly explained? What I did was notice that from the generic form of a vector of S I could form the vectors I wrote in this step. But I not sure if only writing and saying that them spans S I justify effectively that them spans S.

Now I will see if the vectors are linearly independent: Let $r_1,r_2,r_3$ and $r_4$ reals numbers such that

$r_1(-1,1,0) + r_2(0,0,1) + r_3(-i,i,0) + r_4(0,0,i) = (0+0i,0+0i,0+0i,0+0i)$

Performing vector operations: scalar product and vector addition I get

$(-r_1-r_3i,r_1+r_3i,r_2+r_4)=(0+0i,0+0i,0+0i,0+0i)$

Equal the components of the two vectors I get

$-r_1-r_3i=0+0i$ $\iff$ $r_1=0$ and $r_3=0$

$r_1+r_3i=0+0i$ $\iff$ $r_1=0$ and $r_3=0$

$r_2+r_4=0+0i$ $\iff$ $r_2=0$ and $r_4=0$ Are these steps correct? Are they well justified?

$\Longrightarrow$ $r_1=r_3=r_1=r_2=r_4=0$

Therefore the vectors are linearly independent.

Then, since the vectors $(-1,1,0), (0,0,1),(-i,i,0)$ and $(0,0,i)$ are linearly independent and spans S, they form a basis of S.

Is my proof correct?

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    $\begingroup$ I guess you meant $\mathbb{C}^3$, instead of $\mathbb{C}^{3 \times 3}$, as the second one is the vector space of $3\times 3$ matrices with complex entries. On the other hand everything looks OK.. $\endgroup$
    – Stefan4024
    Jun 12 '18 at 20:08
  • $\begingroup$ You are right! thank you for the correction! The demonstration of linear independence and how I got the set of spans is fine? @Stefan4024 $\endgroup$
    – Ayesca
    Jun 12 '18 at 20:13
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    $\begingroup$ Pretty much fine. $\endgroup$
    – Stefan4024
    Jun 12 '18 at 20:18
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The steps are correct, but you're pulling the vectors out of nothing.

You can consider the $\mathbb{R}$-linear map $f(z_1,z_2,z_3)=z_1+z_2$. If you take the bases $\mathscr{B}=\{(1,0,0),(i,0,0),(0,1,0),(0,i,0),(0,0,1),(0,0,i)\}$ of $\mathbb{C}^3$ and $\mathcal{D}=\{1,i\}$ of $\mathbb{C}$, then the matrix of $f$ is $$ \begin{bmatrix} 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \end{bmatrix} $$ A basis of the null space can be computed as $$ \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -1 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ Calling $v_1,\dots,v_6$ the vectors in $\mathscr{B}$, this corresponds $$ \{-v_1+v_3,-v_2+v_4,v_5,v_6\} $$ as a basis for the kernel of $f$.

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  • $\begingroup$ Thanks for the answer and the clarification, @egreg . Is there another way which the vectors are obtained correctly and that they do not seem to come out of nowhere? Because I have not been taught the concepts of null space or kernel $\endgroup$
    – Ayesca
    Jun 19 '18 at 22:56
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    $\begingroup$ @Ayesca Then no. But this is an example of how the concepts of linear map, kernel and matrix representation are useful. $\endgroup$
    – egreg
    Jun 19 '18 at 22:59
  • $\begingroup$ I see, thank you very much $\endgroup$
    – Ayesca
    Jun 19 '18 at 23:02

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