Showing Vector Span Intersections in Fields

[likearollings]

Homework Statement

Let V be a vector over a field F.

a.) Let x1,...,xn∈V and y1,...,ym∈V. Show that

Span(x1,...,xn,y1,...,ym) = Span(x1,...,xn) + Span(y1,...,ym)

B.) Let x1, x2, x3, x4 be four linearly independent vectors in V. Show hat

Span(x1, x2,x3) ∩ Span(x2, x3, x4) = Span(x2,x3)

c.) Show that the equality in part b.) does not hold if we drop the assumption that x1, x2, x3, x4 are linearly independent.

Homework Equations

The Attempt at a Solution

I have done a and b just not sure about c:

If x1=x2=x3=x4 and For a ∈ in R,

Span(x1, x2,x3) ∩ Span(x2, x3, x4)= (a1x1+a2x2+a3x3) ∩ (a2x2+a3x3+a4x4)=a1x1+a2x2+a3x3+a4x4= Span(x1, x2, x3, x4)≠ Span(x2,x3)

Is this okay?

or is it ok to just do a counter example using vectors,

if so could somebody show me an example counter example

thanks in advance

 

[Dick]

No, whatever it is you think you did is not ok. Picking x1=x2=x3=x4 is not going to give you a counter example. Suppose you pick x1=x2=x3 and x4 to be 'something else'. Use numerical vectors to make it super clear.

 

[likearollings]

No, whatever it is you think you did is not ok. Picking x1=x2=x3=x4 is not going to give you a counter example. Suppose you pick x1=x2=x3 and x4 to be 'something else'. Use numerical vectors to make it super clear.

How about x1=x4

then let v∈Span(x1, x2,x3) and v∈Span(x2, x3,x4)

Then

v = a1x1 + a2x2 + a3x3

and v = b1x2 + b2x3 + b3x4

so we have: a1x1 + a2x2 + a3x3 = b1x2 + b2x3 + b3x4

equating coefficents we get

a2=b1
a3=b2

and as x1 = x4:

a1 = b3

Therefore let:

a1 = b3 = c1
a2 = b1 = c2
a3 = b2 = c3


then v = c1x1 + c2x2 + c3x3

so v = (c1/2)x1 + (c1/2)x1 + c2x2 + c3x3

so v = (c1/2)x1 + (c1/2)x4 + c2x2 + c3x3

so v∈Span(x1, x2,x3,x4)



hows that?

 

[likearollings]

No, whatever it is you think you did is not ok. Picking x1=x2=x3=x4 is not going to give you a counter example. Suppose you pick x1=x2=x3 and x4 to be 'something else'. Use numerical vectors to make it super clear.

ok ill try the numerical method tomorow, need some sleep now!

any comments by anyone throughout the night will be massivley appreciated

 

[Dick]

x1=x2=x3=(1,0), x4=(0,1). I'm really starting to wonder if you got a) and b) right.