# base depending on the parameter

Need help $V = \text{lin}((1,1,3,2),(4,5,2,5),(2,3,-4,1),(1,2,-5,5))$

a)Find base and $dimV$

b)For which $t \in \mathbb{R}$, exist base $\alpha1, \alpha2, \alpha3, \alpha4 \in \mathbb{R}^4$ such $\alpha1 \in \mathbb{V}, \alpha1 + 2\alpha3 \in \mathbb{V}$ and vector $\beta = (1,1,3,3)$ in the base has coordinate $(0,t,1,0)$? For each $t$ give an exepmle.

a) base is$(1,1,3,2),(4,5,2,5),(1,2,-5,5)$ and $dimV=3$ right?

b) $\beta \notin V$

$\beta = t\alpha2 + \alpha3$

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@Babak Sorouh there was mistake – aiki93 Jan 31 '13 at 22:03

For a if you set a matrix, say A, whose rows are the vectors in $V$ and then find the corresponding Reduced Row Echelon Form, you will have: $$A=\begin{pmatrix} 1 & 1 & 3 & 2\\ 4 & 5 & 2 & 5\\ 2&3 &-4 &1\\ 1&2 &-5 &5 \end{pmatrix}\longrightarrow \begin{pmatrix} 1 & 0 & 0 & -34\\ 0 & 1 & 0 & 27\\ 0 & 0 & 1 & 3\\ 0&0 &0 &0 \end{pmatrix}$$ This means that the vectors is linearly dependent and the forth vector (as you see in the forth row of the matrix $A$), is a linear combinations of three other vectors. So if $V=\langle v_1,v_2,v_3,v_4\rangle$, then $dim(V)=3$. So your attempts for finding a basis is true in $a$.
sorry in V is $(4,5,2,5)$ not $(1,-1,1,1)$ – aiki93 Jan 31 '13 at 18:50