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Determine whether this is a subspace

  • Thread starter danago
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  • #1
danago
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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.

[tex]
\begin{array}{l}
\left\{ {x \in R^n |Ax = 2x} \right\} \\
\left\{ {x \in R^n |Ax = 2x + e_1 } \right\} \\
\left\{ {x \in R^n |Ax = Bx} \right\} \\
\left\{ {x \in R^n |Ax = Bx + e_1 } \right\} \\
\end{array}
[/tex]



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

For the second and last ones, ive 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 [tex]a,b \in \left\{ {x \in R^n |Ax = 2x} \right\}[/tex], and then show that any linear combination of a and b is also within the set.

[tex]
A(ka + lb) = k(Aa) + l(Ab) = k(2a) + l(2b) = 2(ka + lb)
[/tex]

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.
 

Answers and Replies

  • #2
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your reasoning seems correct.
 
  • #3
HallsofIvy
Science Advisor
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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.
 
  • #4
danago
Gold Member
1,122
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Alright thanks for the confirmation :smile:
 

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