Question

Let V be the solution space of the following homogeneous linear system: x1 − x2 − 2x3 + 2x4 − 3x5 = 0 x1 − x2 − x3 + x4 − 2x5 = 0. (a) (2 points) Find a basis S of V and write down the dimension of V.
(b) (3 points) Finda subspace W of R5 suchthat W contains V anddim(W) = 4. Justify your answer.

Answer 1

$x_1-x_2-2x_3+2x_4-3x_5=0$

$x_1-x_2-x_3+x_4-2x_5=0$

Solving both of them we get;

$x_1-x_2=2x_3-2x_4+3x_5$

$x_1-x_2=x_3-x_4+2x_5$

$2x_3-2x_4+3x_5=x_3-x_4+2x_5$

$x_4=x_3+x_5$

$x_1-x_2=x_3-x_4+2x_5$

$x_1-x_2=x_5$

$x_1=x_2+x_5$

$\begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4\\ x_5 \end{pmatrix}=\begin{pmatrix} x_2+x_5 \\ x_2 \\ x_3\\ x_3+x_5\\ x_5 \end{pmatrix}$

$\begin{pmatrix} x_2+x_5 \\ x_2 \\ x_3\\ x_3+x_5\\ x_5 \end{pmatrix}=\begin{pmatrix} x_2\\ x_2\\ 0\\ 0\\ 0 \end{pmatrix}+\begin{pmatrix} x_5\\ 0\\ 0\\ x_5\\ x_5 \end{pmatrix}+\begin{pmatrix} 0\\ 0\\ x_3\\ x_3\\ 0 \end{pmatrix}$

$\begin{pmatrix} x_2+x_5 \\ x_2 \\ x_3\\ x_3+x_5\\ x_5 \end{pmatrix}=\begin{pmatrix} 1\\ 1\\ 0\\ 0\\ 0 \end{pmatrix}x_2+\begin{pmatrix} 1\\ 0\\ 0\\ 1\\ 1 \end{pmatrix}x_5+\begin{pmatrix} 0\\ 0\\ 1\\ 1\\ 0 \end{pmatrix}x_3$

${\begin{pmatrix} 1\\ 1\\ 0\\ 0\\ 0 \end{pmatrix},\begin{pmatrix} 1\\ 0\\ 0\\ 1\\ 1 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1\\ 1\\ 0 \end{pmatrix}}$ Is the basis of V.

Dimension of V is 3.

(b)If W is a solution space of homogenous system of equation $x_1-x_2-2x_3+2x_4-3x_5=0$

$x_1=x_2+2x_3-2x_4+3x_5$

$\begin{pmatrix} x_1 \\ x_2 \\ x_3\\ x_4\\ x_5 \end{pmatrix}=\begin{pmatrix} x_2+2x_3-2x_4+3x_5 \\ x_2 \\ x_3\\ x_4\\ x_5 \end{pmatrix}$

$\begin{pmatrix} x_2+2x_3-2x_4+3x_5 \\ x_2 \\ x_3\\ x_4\\ x_5 \end{pmatrix}=\begin{pmatrix} x_2\\ x_2\\ 0\\ 0\\ 0 \end{pmatrix}+\begin{pmatrix} 2 x_3\\ 0\\ x_3\\ 0\\ 0 \end{pmatrix}+\begin{pmatrix} -2x_4\\ 0\\ 0\\ x_4\\ 0 \end{pmatrix}+\begin{pmatrix} 3x_5\\ 0\\ 0\\ 0\\ x_5 \end{pmatrix}$

$\begin{pmatrix} x_2+2x_3-2x_4+3x_5 \\ x_2 \\ x_3\\ x_4\\ x_5 \end{pmatrix}=\begin{pmatrix} 1\\ 1\\ 0\\ 0\\ 0 \end{pmatrix}x_2+\begin{pmatrix} 2 \\ 0\\ 1\\ 0\\ 0 \end{pmatrix}x_3+\begin{pmatrix} -2\\ 0\\ 0\\ 1\\ 0 \end{pmatrix}x_4+\begin{pmatrix} 3\\ 0\\ 0\\ 0\\ 1 \end{pmatrix}x_5$

${\begin{pmatrix} 1\\ 1\\ 0\\ 0\\ 0 \end{pmatrix},\begin{pmatrix} 2 \\ 0\\ 1\\ 0\\ 0 \end{pmatrix} ,\begin{pmatrix} - 2 \\ 0\\ 0\\ 1\\ 0 \end{pmatrix},\begin{pmatrix} 3 \\ 0\\ 0\\ 0\\ 1 \end{pmatrix}}$ Is the basis of W.

Dimension of vector space W is 4.