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In linear space of matrix $2\times 3$ over $C$ we have subspace generated by: $ A= \{{\left[\begin{array}{ccc}i&i&i\\i&0&1\end{array}\right]}$ ${\left[\begin{array}{ccc}i&i&-i\\-i&-1&-1\end{array}\right]}$ ${\left[\begin{array}{ccc}2&2&0\\0&-i&0\end{array}\right]}$ ${\left[\begin{array}{ccc}i&-i&-i\\0&0&2\end{array}\right]} \}$

From vectors in set A choose basis $span(A)$

I'm not sure if it's good but I would try to type matrices to column like this: ${\left[\begin{array}{ccc}i&i&2&i\\i&i&2&-i\\i&-i&0&i\\i&-i&0&0\\0&-1&i&0\\1&-1&0&2\end{array}\right]}$ and reduce it and convert it to $4$ matrices at the end, is my idea ok ?

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That's exactly the right idea...but type the matrices as ROWS and then ROW-reduce (or you could column-reduce). You're starting with four matrices, so you expect to end up with no more than 4 independent rows.

Looking at it for 5 seconds, it looks to me as if the resulting basis will have either 3 or 4 matrices.

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  • $\begingroup$ Ok it will looks like this so the answer is $4$ matrices with this coordinates ? $\endgroup$ – Mario Nov 24 '14 at 12:23
  • $\begingroup$ That's what I got (doing it by hand), yes. $\endgroup$ – John Hughes Nov 25 '14 at 13:15

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