Subspace Span Determination for Vector y in R^4 using Augmented Matrix

[karnten07]

Homework Statement

For each s \inR determine whether the vector y is in the subspace of R^4 spanned by the columns of A where

y=
6
7
1
s

and A =

1 3 2
-1 -1 1
3 8 1
4 9 3

Homework Equations

The Attempt at a Solution

Can i do this by making an augmented matrix from A and y and then reducing it, then i can see what value of s is required? i will attemp this and update with what i get soon.

 

[karnten07]

Okay, so i get my reduced row form as:

1 3 2 6
0 1 3 13
0 0 1 2
0 0 0 s+7

so does this tell me that s must be -7?

So would i say that for s = -7, the vector y is in the subspace of R^4 and for s \neq-7 it is not in the subspace? Is this all that is required? Do i need to show that the vector y with s =-7 is closed under additon and scalar multiplication, if so i may need help showing this. Many thanks

 

[HallsofIvy]

Yes, that's all you need to show.

You don't need "to show that the vector y with s =-7 is closed under additon and scalar multiplication"- that doesn't even make sense! A vector is never closed under addition and scalar multiplication- a subspace is. An the "span" of a set of vectors, including the three vectors (not y) given here is, by definition, all possible sums of scalar products of the vectors and so is closed under addition and scalar multiplication.

You could have done this directly from the definition of "span": y is in the span if and only if it is equal to a linear combination of them: that (6, 7 , 1, s)= a(1, -1, 3, 4)+ b(3, -1, 8, 9)+ c(2, 1, 1, 3) for some values of a, b, c. Of course, that gives you 4 equations to solve: a+ 3b+ 2c= 6, -a- b+ c= 7, 3a+ 8b+ 3c= 1, and 4a+ 9b+ 3c= s and solving that system is exactly the same as reducing the augmented matrix you have.

 

[karnten07]

Yes, that's all you need to show.

You don't need "to show that the vector y with s =-7 is closed under additon and scalar multiplication"- that doesn't even make sense! A vector is never closed under addition and scalar multiplication- a subspace is. An the "span" of a set of vectors, including the three vectors (not y) given here is, by definition, all possible sums of scalar products of the vectors and so is closed under addition and scalar multiplication.

You could have done this directly from the definition of "span": y is in the span if and only if it is equal to a linear combination of them: that (6, 7 , 1, s)= a(1, -1, 3, 4)+ b(3, -1, 8, 9)+ c(2, 1, 1, 3) for some values of a, b, c. Of course, that gives you 4 equations to solve: a+ 3b+ 2c= 6, -a- b+ c= 7, 3a+ 8b+ 3c= 1, and 4a+ 9b+ 3c= s and solving that system is exactly the same as reducing the augmented matrix you have.

Thats great, thanks HallsofIvy.

 

[squenshl]

I have a problem.
Suppose that {u1,u2,...,um} are vectors of R^n. Prove, directly that span
{u1,u2,...,um} is a subspace of R^n.

 

[HallsofIvy]

I have a problem.
Suppose that {u1,u2,...,um} are vectors of R^n. Prove, directly that span
{u1,u2,...,um} is a subspace of R^n.

Since this is a separate problem, start a separate thread. The "New Topic" button is just above the list of threads.