Proving Subspaces of Vector Spaces: Evaluating A Vector x

[Supernova123]

Homework Statement

How would one determine if a vector space is a subspace of another one? I think that the basis vectors of the subspace should be able to be formed from a linear combination of the basis vectors of the vector space.

However, that doesn't seem to be true for this question: Let matrix A consist of column vectors (1,2,-3), (-4,-4,4) and (6,2,-8) with eigenvalues -2 and -5. I found e1=(2,3,1) and e2=(1,0,-1).

The linear space spanned by e1 and e2 is denoted by V. Prove that, for any vector x belonging to V, the vector Ax also belongs to V. I tried forming any of the column vectors in matrix A through a linear combination of e1 and e2 but that seems to fail. Instead, the answer is this: Let x=ae1+be2. Ax=A(ae1+be2)=aAe1+bAe2=-2ae1-5ae2. I understand this but why doesn't my method work?

Homework Equations

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

How would one determine if a vector space is a subspace of another one? I think that the basis vectors of the subspace should be able to be formed from a linear combination of the basis vectors of the vector space.

However, that doesn't seem to be true for this question: Let matrix A consist of column vectors (1,2,-3), (-4,-4,4) and (6,2,-8) with eigenvalues -2 and -5. I found e1=(2,3,1) and e2=(1,0,-1).

The linear space spanned by e1 and e2 is denoted by V. Prove that, for any vector x belonging to V, the vector Ax also belongs to V. I tried forming any of the column vectors in matrix A through a linear combination of e1 and e2 but that seems to fail. Instead, the answer is this: Let x=ae1+be2. Ax=A(ae1+be2)=aAe1+bAe2=-2ae1-5ae2. I understand this but why doesn't my method work?

Homework Equations

The Attempt at a Solution

Isn't Ax = -2ae1-5be2 a linear combination of the vectors e1 and e2?

 

[Supernova123]

Ax=a(1,2,-3)+b(-4,-4,4)+c(6,2,-8) and V contains Ax. So wouldn't de1+fe2=(1,2,-3) or (-4,-4,4) or (6,2,-8)?

 

[Ray Vickson]

Ax=a(1,2,-3)+b(-4,-4,4)+c(6,2,-8) and V contains Ax. So wouldn't de1+fe2=(1,2,-3) or (-4,-4,4) or (6,2,-8)?

Think about it: as x ranges over S = R³ (the whole space), Ax ranges over S as well (because for any y in S the equation Ax = y has a unique solution). However, your defined subspace V (= set of all linear combinations of e1 and e2) is only two-dimensional, so cannot contain all three rows of A. In other words, a two-dimensional space is not a three-dimensional space.