Show that the vectors $v_1$ = (i,1+i,2+i), $v_2$ = (1,1+i,2+i), and $v_3$ = (2,-i,-i) form a basis for the complex vector space $C^3$....

Show that $v_1,iv_1,v_2,iv_2,v_3,iv_3$ is a basis of $C^3$ as a vector space over R. What are complex and real dimensions of this space?

Don't worry i think i did the First couple of parts. In order for the vectors to be the basis for $C^3$ the system they make must be liinearly independent span $C^3$ $$av_1+bv_2+cv_3=0$$ meaning... $$a=b=c=0$$

Working with matrix $$\begin{bmatrix} i&1&2 \\ 1+i&1+i&-i \\ 2+i&2+i&-i \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix}$$

which i row reduced to... $$\begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 0\\0\\0 \end{bmatrix}$$

Since a=b=c=0 the {a,b,c} is linearly independent and spans $\mathbb C^3$ as it has 3 pivots. This also means vectors v1,v2 and v3 are a basis of $\mathbb C^3$ as a vector space over $\mathbb R$.

I guess the complex dimensions are complex elements of the vectors such $iv_3$ = {0,-i,-i} and the real dimensions is simply the real component i.e. $v_3$ = {2,0,0}

  • $\begingroup$ "This also means vectors $v_1$, $v_2$ and $v_3$ are a basis of $\mathbb C^3$ as a vector space over $\mathbb R$."?? - At any rate, ultimately dimensions should be numbers. $\endgroup$ – Hagen von Eitzen Aug 18 '15 at 14:36
  • $\begingroup$ Yeah i didn't quite understand the bit on dimensions. $\endgroup$ – Ivan Aug 18 '15 at 15:53
  • $\begingroup$ Hey can anyone explain the bit about complex dimensions to me? $\endgroup$ – Ivan Aug 18 '15 at 22:52

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