# Finding the basis and dimension of a vector space

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 $(-2,-12,-4,-20), (0,70,14,98)$ ?

• Where are $a_4$ and $b_4$? – NalRa Dec 25 '14 at 14:03
• Given only these. – I.A. G Dec 25 '14 at 17:37
• @I.A.G read the question carefully. Are you sure that there is supposed to be an $a_4$ and a $b_4$? – Omnomnomnom Dec 25 '14 at 17:48
• @Omnomnomnom,Probably, it's mistake. – I.A. G Dec 25 '14 at 18:08

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 $\{a_1,a_2\}$.
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 $L_1 \cap L_2$.
• Very nice, efficient way to answer the $L_1+L_2$ and $L_1\cap L_2$ questions. – JohnD Dec 25 '14 at 18:17