Determine whether this is a subspace

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In summary, the first and third subsets of Rn (where Ax=2x and Ax=Bx, respectively) are subspaces since they are closed under addition and scalar multiplication. The second and fourth subsets (where Ax=2x+e1 and Ax=Bx+e1, respectively) are not subspaces since the zero vector does not satisfy the equations and therefore is not an element of the sets.
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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 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 [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.
 
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  • #2


your reasoning seems correct.
 
  • #3


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


Alright thanks for the confirmation :smile:
 

FAQ: Determine whether this is a subspace

What is a subspace?

A subspace is a subset of a vector space that satisfies the properties of a vector space. These properties include closure under vector addition and scalar multiplication, and containing the zero vector.

What are the properties of a subspace?

The properties of a subspace include closure under vector addition and scalar multiplication, and containing the zero vector. In other words, for any two vectors in the subspace, their sum must also be in the subspace, and any scalar multiple of a vector in the subspace must also be in the subspace.

How do you determine if something is a subspace?

To determine if something is a subspace, you must check if it satisfies the properties of a subspace. This means checking if the subset is closed under vector addition and scalar multiplication, and if it contains the zero vector. If all three properties are satisfied, then the subset is a subspace.

What is the importance of subspaces?

Subspaces are important in linear algebra because they provide a way to study and analyze vector spaces in a more simplified manner. By focusing on subspaces, we can better understand the structure and properties of a vector space as a whole.

What are some examples of subspaces?

Some examples of subspaces include the set of all 2x2 matrices, the set of all polynomials of degree 3 or less, and the set of all real-valued functions on a given interval. These subsets satisfy the properties of a subspace and are therefore considered subspaces of their respective vector spaces.

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