Determine whether this is a subspace

[danago]

Let A,B be n x n matrices and e₁ be the first standard basis vector in Rⁿ. For each of the following subsets of Rⁿ, determine whether it is a subspace of Rⁿ, giving reasons.

<br /> \begin{array}{l}<br /> \left\{ {x \in R^n |Ax = 2x} \right\} \\ <br /> \left\{ {x \in R^n |Ax = 2x + e_1 } \right\} \\ <br /> \left\{ {x \in R^n |Ax = Bx} \right\} \\ <br /> \left\{ {x \in R^n |Ax = Bx + e_1 } \right\} \\ <br /> \end{array}<br />



Now i think I am ok with the second and last ones, but not so sure about the other two.

For the second and last ones, I've noticed that the zero vector does not satisfy the equation, thus the zero vector is not within the set and hence the set is not a subspace.

For the first and third, the zero vector is an element of the sets, which is one of the conditions of a subspace. The other condition would be that the set is closed under both addition and multiplication.

What i considered was to let a,b \in \left\{ {x \in R^n |Ax = 2x} \right\}, and then show that any linear combination of a and b is also within the set.

<br /> A(ka + lb) = k(Aa) + l(Ab) = k(2a) + l(2b) = 2(ka + lb)<br />

Where k and l are arbitary real constants.

And according to my understanding, this shows that any linear combination of a and b are within the set, hence the set is a subspace?

Is my reasoning correct? Id then apply the same reasoning to the third set.

Thanks,
Dan.

 

[foxjwill]

your reasoning seems correct.

 

[HallsofIvy]

Yes, in general one shows that a subset of a vector space is a subspace by showing that the set is closed under addition and scalar multiplication- which is the same as saying that any linear combination, ka+ lb, where k and l are scalars and a and b are vectors in the subset, is also in that subset.

Because A is a matrix (more generally, a linear transformation), A(ka+ lb)= k(Aa)+ l(Ab). Since a and b are in the subset, Aa= 2a, Ab= 2b so A(ka+ lb)= k(2a)+ l(2b)= 2(ka+ lb), showing that ka+ lb also satisfies that equation and so is in that subset.

 

[danago]

Alright thanks for the confirmation