Linear algebra question

Question

Show that the following sets of vectors of R3 are linearly independent

and span R3:

(a) S1 = {(1, 2, 3),(3, 2, 1),(0, 0, 1)}.

(b) S2 = {(3, 0, −6),(−4, 1, 7),(−2, 1, 5)}.

Answer 1

Hi Osama,

In order for a set of vectors to be linearly independent, they need to satisfy the condition that if you add up all three vectors with a coefficient in front of each, then the all the coefficients must be zero for it to be linearly independent. Furthermore, a set spans the space if there are n basis elements for an n dimensional space. In other words, if I chose a random three dimensional vector and set it equal to the sum of all the linearly independent vectors I just found, then we would say the set of vectors spans the space if I can find an expression with unique coefficients for the linearly independent vectors that is equal to the variable three dimensional vector.

So, for example, the set (1,2,3) and (1,2,4) are linearly independent, however they do not span the entire space. Namely, they span only one subspace that of (1,2,z). In other words, I could never find a vector (2,3,5) no matter how cleverly, I chose my coefficients.

On the other hand, the set of vectors (1,2,3), (1,0,2), (2,3,4) are linearly independent and span the entire space. I will leave checking the linear independence and show you why they span:

a(1,2,3)+b(1,0,2)+c(2,3,4)=(e,f,g) where e,f,g are real numbers.

(a+b+2c,2a+3c,3a+2b+4c)=(e,f,g)

Which gives us the following equations:

(1) a+b+2c=e

(2) 2a+3c=f

(3) 3a+2b+4c=g

(4) Subtracting 2(1)-(2) => b+4c=2e-f

(5) Subtracting (3)-2(1) => 2a=g-2e=> a=(g-2e)/2

Substituting a in (2) => 3c=f-2(g-2e)/2=> c=(f-g+2e)/3

Substituting c in (4) => b=2e-f-4(f-g+2e)/3

Thus, we have found nonzero values for a,b,c that will give us any random three dimensional vector. Therefore, we say that it spans all of R³ .

Now that I have you started, just apply this process to your own questions.