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Find the basis and dimension of vector space $ L_{1}$ spanned by vectors $ a_{1} ,a_{2},a_{3} $, the basis and dimension of vector space $ L_{2}$ spanned by vectors $ b_{1} ,b_{2},b_{3} $ and also $ L_{1} + L_{2} , L_{1} \cap L_{2}$ $ a_{1} = (-10,10,-6,-2) ,a_{2} = (-6,4,-4,-4) ,a_{3} = (-2,-12,-4,-20) $ $ b_{1} = (6,3,-6,4) ,b_{2} = (-4,0,0,-3) ,b_{3} = (-13,-7,9,-12) $

I've tried this: \begin{pmatrix}-10 & 10 & -6 & -2\\ -6 & 4 & -4 & -4\\ -2 & -12 & -4 & -20 \end{pmatrix}

\begin{pmatrix}-2 & -12 & -4 & -20\\ 0 & 70 & 14 & 98\\ 0 & 0 & 0 & 0 \end{pmatrix}

So basis for $ L_{1} $ is $ (-2,-12,-4,-20), (0,70,14,98) $ ?

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    $\begingroup$ Where are $a_4$ and $b_4$? $\endgroup$ – NalRa Dec 25 '14 at 14:03
  • $\begingroup$ Given only these. $\endgroup$ – I.A. G Dec 25 '14 at 17:37
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    $\begingroup$ @I.A.G read the question carefully. Are you sure that there is supposed to be an $a_4$ and a $b_4$? $\endgroup$ – Omnomnomnom Dec 25 '14 at 17:48
  • $\begingroup$ @Omnomnomnom,Probably, it's mistake. $\endgroup$ – I.A. G Dec 25 '14 at 18:08
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For the first, form a matrix $A$ whose columns are the vectors $a_i$ and row reduce it: $$\begin{bmatrix} -10 & -6 & -2 \\ 10 & 4 & -12 \\ -6 & -4 & -4 \\ -2 & -4 & -20 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & -4 \\ 0 & 1 & 7 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $$

There are leading ones in the first and second column of the row reduced matrix, so that means the first and second columns of $A$ (i.e., the vectors $a_1$ and $a_2$) are linearly independent and span the space $\text{span}\{a_1,a_2,a_3\}$. Thus $L_1$ is two dimensional and a basis for $L_1$ is $\{a_1,a_2\}$.

Adapt this technique to answer your other questions.

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Your method is certainly a correct way of obtaining a basis for $L_1$. You can then do the same for $L_2$. Another method is that outlined by JohnD in his answer.

Here's a neat way to do the rest, analogous to this second method: suppose that $u_1,u_2$ is a basis of $L_1$, and that $v_1,v_2,v_3$ (there may be no $v_3$) is a basis of $L_2$. Row reduce the matrix with columns $$ \pmatrix{u_1&u_2&v_1&v_2&v_3} $$ The columns corresponding to the resulting pivots form a basis of $L_1 + L_2$. The other columns correspond to a basis of $L_1 \cap L_2$.

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  • $\begingroup$ Very nice, efficient way to answer the $L_1+L_2$ and $L_1\cap L_2$ questions. $\endgroup$ – JohnD Dec 25 '14 at 18:17

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