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Let $$E_k = \begin{pmatrix} 2 \\ 1 \\ k \\ -1 \end{pmatrix} + Span \left( \begin{pmatrix} 2+k \\ 3 \\ 2k \\ 2 \end{pmatrix}, \begin{pmatrix} -2 \\ k-5 \\ 4 \\ k-3 \end{pmatrix} \right).$$ Find $n_1, n_2 \in \mathbb{N}$ and $k_0 \in \mathbb{R}$ such that: (a) $\dim(E_k)=n_1$ for all $k \neq k_0$; (b) $\dim(E_{k_0})=n_2$.

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$\dim(E_{k})=2$ if $\begin{vmatrix} 2+k & -2 \\ 3 & k-5 \end{vmatrix} \neq 0 \Leftrightarrow k \neq 4 \land k \neq -1$ –  Katy23 May 29 '11 at 9:56
    
the solution of this problem are $n_1=2$, $n_2=1$ and $k_0=-1$. But why? –  Katy23 May 29 '11 at 9:59
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Question's a little odd. With that $(2,1,k,1)$ out front, $E_k$ is not a vector space, so what does dimension mean? I'll take it to mean some kind of affine dimension, but it seems to me that the $(2,1,k,1)$ is a red herring, so I'll make believe it's not there.

Then as girdav notes the question is, what's the dimension of the span. It will be 2 unless one vector is a multiple of the other. This requires the existence of a number $c$ such that $2+k=-2c$, $3=(k-5)c$, $2k=4c$, and $2=(k-3)c$. Can you find the values of $c$ and $k$, if any, that satisfy all 4 equations?

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