# Thread: Linear combination is also in the span

1. ## Linear combination is also in the span

Suppose that vectors x and y are both in the linear span of the
vectors r1, r2, r3. Show that any linear combination of x and y is also
in the linear span of r1, r2, r3.

What I have-

Because x and y are in the span of r1,r2,r3, then x and y are linear combinations of r1,r2,r3. So, the following exists:

c1(r1)+c2(r2)+c3(r3) = x

and

d1(r1)+d2(r2)+d3(r3) = y

where the d's and c's are scalars. I'm not sure this gets me anywhere with the proof.

Thanks.

2. ## Re: Linear combination is also in the span

if:

$x = c_1r_1 + c_2r_2 + c_3r_3$, and $y = d_1r_1 + d_2r_2 + d_3r_3$,

then $ax+by = a(c_1r_1 + c_2r_2 + c_3r_3) + b(d_1r_1 + d_2r_2 + d_3r_3)$

$= (ac_1)r_1 + (ac_2)r_2 + (ac_3)r_3 + (bd_1)r_1 + (bd_2)r_2 + (bd_3)r_3$

$= (ac_1+bd_1)r_1 + (ac_2+bd_2)r_2 + (ac_3+bd_3)r_3$ which is surely in span({r1,r2,r3})

since for each i, $ac_i + bd_i$ is surely some scalar.

3. ## Re: Linear combination is also in the span

Originally Posted by sfspitfire23
Suppose that vectors x and y are both in the linear span of the vectors r1, r2, r3. Show that any linear combination of x and y is also
in the linear span of r1, r2, r3.
Suppose that $x = \chi _1 r_1 + \chi _2 r_2 + \chi _3 r_3 \;\& \,y = \gamma _1 r_1 + \gamma _2 r_2 + \gamma _3 r_3$.

Show that $\alpha x + \beta \,y = \omega _1 r_1 + \omega _2 r_2 + \omega _3 r_3$.

i.e. $\omega _k=~?$