Let W x R5 2x1 x3 x5 0 x2 3x4 x5 0 x1 x2 x4 2x5 0 Sho

Let W ={x R⁵: 2x₁ + x₃ x₅ = 0, x₂ + 3x₄ + x₅ = 0, x₁ + x₂ x₄ 2x₅ = 0. Show that W is a subspace of R⁵, and then find a spanning set for W.

0 vector ie (0,0,0,0,0) belogns to W

Let, (a1,a2,a3,a4,a5) and (b1,b2,b3,b4,b5) belong to W

So,

Sum of vectors is (a1+b1,a2+b2,a3+b3,a4+b4,a5+b5)

2(a1+b1)+a3+b3-(a5+b5)=2a1+a3-a5+2b1+b3-b5=0

(a2+b2)+3(a4+b4)+(a5+b5)=a2+3a4+a5+b2+3b4+b5=0

(a1+b1)+a2+b2 -(a4+b4)-2(a5+b5)=0

Sum sum of any two vectors in W is also in W

Hence, W is a subspace of R5

There are 5 variables and 3 equations. So we will have 2 free variables.

LEt us choose x5 and x4 as the free variables

So substituting in second equation gives

x2=-3x4-x5

Substituting in third equation gives

x1+(-3x4-x5)-x4-2x5=0

x1-4x4-3x5=0

x1=4x4+3x5

SUbstituting in first equation gives

2x1+x3-x5=0

2(4x4+3x5)+x3-x5=0

x3=-8x4-5x5

Hence general solution is

(4x4+3x5,-3x4-x5,-8x4-5x5,x4,x5)=x4(4,-3,-8,1,0)+x5(3,-1,-5,0,1)

Hence spanning set is

{(4,-3,-8,1,0),(3,-1,-5,0,1)}