Answer to Question #254029 in Linear Algebra for usama

Question

Show that the vectors u1 = (1, 1, 1), u2 = (1, 2, 3), u3 = (1, 5, 8) span R

3

.

Answer 1

Solution:

To solve it show that an arbitrary vectorsl v=(a,b,c) $\in$ R³ is a linear combination of $u_1,u_2$ and $u_3$ .

Let

$v=xu_1+yu_2+zu_3$

Hence,

$(a,b,c)=x(1,1,1)+y(1,2,3)+z(1,5,8)$

Resolve as

$x+y+z=a$

$x+2y+5z=b$

$x+3y+8z=c$

Form the equivalent matrix form of the equations

$\begin{bmatrix} 1&1& 1&\vert&a \\ 1&2 & 5&\vert&b\\ 1&3&8&\vert&c \end{bmatrix}$

Reduced echelon form is

$\begin{bmatrix} 1 & 1&1&\vert&a\\ 0 & 1&1&\vert&b-a\\ 0&0&-1&\vert&a-2b+c \end{bmatrix}$

From the reduced echelon form,

$-z=a-2b+c$

$z=-a+2b-c$

$y+4z=b-a$

$y=b-a-4z$

By substitution,

$y=b-a-4(-a+2b-c)$

$y=3a-7b+4c$

Also from the echelon form,

$x+y+z=a$

$x=a-y-z$

By substitution,

$x=a-(3a-7b+4c)-(-a+2b-c)$

$x=-a+5b-3c$

Therefore,

$(a,b,c)=(-a+5b-3c)(1,1,1)+(3a-7b+4c)(1,2,3)+(-a+2b-c)(1,5,8)$

Therefore,

$u_1,u_2,u_3$ span $R^3$