(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that the set:

S = {x[itex]\in[/itex] R[itex]^{4}[/itex]|x= [itex]\lambda[/itex](2,0,1,-1)[itex]^{T}[/itex] for some [itex]\lambda[/itex] [itex]\in[/itex] R

is a subspace of R[itex]^{4}[/itex]

3. The attempt at a solution

For the subspace theorem to hold, 3 conditions must be met:

1) The zero vector must exist

2) Closed under addition

3) Closed under scalar multiplication

1) If [itex]\lambda[/itex] = 0, the vector becomes (0,0,0,0)[itex]^{T}[/itex] - therefore that's the zero vector.

2) Closure under addition is what I'm a bit confused about.

If we define two new vectors,uandvand two scalars [itex]\alpha[/itex] and [itex]\beta[/itex] respectively.

u+v= ?

3) For closure under multiplication, isn't this obviously already closed? Heck it's being multiplied by a scalar quantity already.

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# Proving a set is a subspace

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