A clarification in linear algebra

[ehabmozart]

Homework Statement

This is a general question in linear algebra

Determine whether the following subset of R^3 is a subspace

The elements go vertically but I can't show them in this way and will show them horizontally however.

( s-2t, s, t+s ) / s, t ε R ...

Homework Equations

The book attempt to show closure under addition and multiplication

The Attempt at a Solution

Is it possible to use another argument. I claim that these three elements can be divided into two vectors which are also subsets of R-3 ... They are the span of (1,1,1)(-2,0,1).. Since they are lin. independent, they are vector spaces?

 

[tiny-tim]

hi ehabmozart!

I claim that these three elements can be divided into two vectors which are also subsets of R-3 ... They are the span of (1,1,1)(-2,0,1).. Since they are lin. independent, they are vector spaces?

you're saying that anything of the form ( s-2t, s, t+s ) is s(1,1,1) + t(-2,0,1) and vice versa

so yes, if you've already proved the relevant theorems (which if you're still on page 1 of the chapter, you haven't!), that does immediately prove they are a vector subspace

 

[PeroK]

Hi, you might want to think about it this way:

Suppose you have the set:

$(a_1s + b_1t, a_2s + b_2t, a_3s + b_3t): s, t \in R$

Where the a's and b's are any constants. Is this set a subspace? Try showing closure under addition and scalar multiplication.

Then, you might want to consider a mapping, L, from $R^2 to R^3$

$L(s, t) = (a_1s + b_1t, a_2s + b_2t, a_3s + b_3t)$

Then, you might want to consider what sort of mapping this is and how L could be represented.

Then, you might look at your original problem in new light.

Interesting? Or, not?

 

[ehabmozart]

hi ehabmozart!


you're saying that anything of the form ( s-2t, s, t+s ) is s(1,1,1) + t(-2,0,1) and vice versa

so yes, if you've already proved the relevant theorems (which if you're still on page 1 of the chapter, you haven't!), that does immediately prove they are a vector subspace

Excuse me, what are the relevant theorems you are talking about?

 

[ehabmozart]

Hi, you might want to think about it this way:

Suppose you have the set:

$(a_1s + b_1t, a_2s + b_2t, a_3s + b_3t): s, t \in R$

Where the a's and b's are any constants. Is this set a subspace? Try showing closure under addition and scalar multiplication.

Then, you might want to consider a mapping, L, from $R^2 to R^3$

$L(s, t) = (a_1s + b_1t, a_2s + b_2t, a_3s + b_3t)$

Then, you might want to consider what sort of mapping this is and how L could be represented.

Then, you might look at your original problem in new light.

Interesting? Or, not?

I am sorry but I don't your point here... Please clarify!

 

[PeroK]

I assume you've done matrices? If so, you might notice that the original set can be seen as the action of the matrix:

$\begin{pmatrix} 1&-2\\ 1&0\\ 1&1\\ \end{pmatrix}$

On vectors (s, t). And, those columns should look familiar!

 

[tiny-tim]

Excuse me, what are the relevant theorems you are talking about?

direct sum of two linear spaces is a linear space?

 

[HallsofIvy]

Homework Statement

This is a general question in linear algebra

Determine whether the following subset of R^3 is a subspace

The elements go vertically but I can't show them in this way and will show them horizontally however.

( s-2t, s, t+s ) / s, t ε R ...

Homework Equations

The book attempt to show closure under addition and multiplication

The Attempt at a Solution

Is it possible to use another argument. I claim that these three elements can be divided into two vectors which are also subsets of R-3 ... They are the span of (1,1,1)(-2,0,1).. Since they are lin. independent, they are vector spaces?

Yes. What your are saying is the set of all vectors of the form (s- 2t, s, t+s)= s(1, 1, 1)+ t(-2, 0, 1) is the span of vectors (1, 1, 1) and (-2, 0, 1) and so is a vector space.

(The span of any set of vectors is a vector space. The fact that those two vectors are independent tells you that this is a two dimensional subspace of $R^3$ and that {(1, 1, 1) (-2, 0, 1)} is a basis.)