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) $\dim(E_{k_0})=n_2$.
A dimension problem
1
$\begingroup$
linear-algebra
-
0$\dim(E_{k})=2$ if $\begin{vmatrix} 2+k & -2 \\ 3 & k-5 \end{vmatrix} \neq 0 \Leftrightarrow k \neq 4 \land k \neq -1$ – 2011-05-29
-
0the solution of this problem are $n_1=2$, $n_2=1$ and $k_0=-1$. But why? – 2011-05-29