Question

Which of the following subsets of R³ are actually subspaces? (a) The plane of vectors (b1,b2,b3) with b1=b2 (b) The plane of vectors with b1=1 (c) The vectors with b1b2b3=0 .

Answer 1

Among the options, only (a) the plane of vectors
`\langle b_(1),\, b_(2),\, b_(3)\rangle` where
`b_(1) = b_(2)` is a subspace.

Step-by-step explanation:

A subset
`W` of a vector space is a linear subspace if and only if both of the following are satisfied:

• The subset
  `W` is closed under vector addition. In other words, if vectors
  `u` and
  `v` are members of
  `W\!`, their sum
  `(\mathbf{u} + \mathbf{v})` should also be a member of
  `W\!\!`:
  `\mathbf{u},\, \mathbf{v} \in W \implies (\mathbf{u} + \mathbf{v}) \in W`.
• The subset
  `W` is closed under scalar multiplication. In other words, for any scalar
  `c \in \mathbb{R}` (a real number) and for any member
  `\mathbf{v}` of
  `W\!`, the scalar multiple
  `c\, \mathbf{v}\!` should also be a member of
  `W\!\!`:
  `(c \in \mathbb{R}) \land (\mathbf{v} \in W) \implies c\, \mathbf{v} \in W`.

To prove that a subset of a vector space is a linear subspace, simply prove that this subset satisfy the two properties above. A counterexample for one of the two properties is sufficient to show that the subset isn't a linear subspace.

(a)

Let
`W` denote the plane of vectors
`\langle b_(1),\, b_(2),\, b_(3)\rangle` where
`b_(1) = b_(2)`. A member of
`\mathbb{R}^(3)` is in
`W\!` if and only if its first two components are equal.

To prove that this
`W\!` is closed under addition, consider members of this subspace
`\mathbf{u},\, \mathbf{v} \in W`. The goal is to show that
`\mathbf{u} + \mathbf{v} \in W` by showing that its first two components are equal.

Let
`\mathbf{u} = \langle u_(1),\, u_(2),\, u_(3) \rangle` and let
`\mathbf{v} = \langle v_(1),\, v_(2),\, v_(3) \rangle` for scalars
`u_(1),\, u_(2),\, u_(3),\, v_(1),\, v_(2),\, v_(3)`.

`\mathbf{u} + \mathbf{v} = \langle u_(1) + v_(1),\, u_(2) + v_(2),\, u_(3)+ v_(3) \rangle`

By the construction of
`W`,
`u_(1) = u_(2)` and
`v_(1) = v_(2)`.. Add the two equalities to obtain:
`u_(1) + v_(1) = u_(2) + v_(2)`. In other words, the first two components of
`(\mathbf{u} + \mathbf{v})` are indeed equal, and
`(\mathbf{u} + \mathbf{v})\!` is a member of
`W`.

To prove that
`W` is closed under scalar multiplication, consider one member of this subspace,
`\mathbf{v} \in W` where
`\mathbf{v} = \langle v_(1),\, v_(2),\, v_(3)\rangle` for scalars
`v_(1),\, v_(2),\, v_(3)`. Let
`c \in \mathbb{R}` be a scalar. The goal is to show that
`c\, \mathbf{v} \in W` by showing that the first two components of
`(c\, \mathbf{v})` are equal.

`c\, \mathbf{v} = \langle c\, v_(1),\, c\, v_(2),\, c\, v_(3)\rangle`.

Since
`\mathbf{v} \in W`,
`v_(1) = v_(2)`, such that
`c\, v_(1) = c\, v_(2)`. Hence,
`c\, \mathbf{v}` is indeed a member of
`W`.

Since
`W \in \mathbb{R}^(3)` satisfies both properties of linear subspace,
`W\!` would indeed be a linear subspace of
`\mathbb{R}^(3)`.

(b)

The following counterexample demonstrates that the plane of vectors
`\langle b_(1),\, b_(2),\, b_(3)\rangle \in \mathbb{R}^(3)` with
`b_(1) = 1` isn't closed under scalar multiplication.

Consider scalar
`c = 0` and vector
`\mathbf{v} = \langle 1,\, 0,\, 0\rangle`. Since the first component of
`\mathbf{v}` is
`1`, this vector would be a member of the given subset. However, the scalar multiple
`c\, \mathbf{v} = \langle 0,\, 0,\, 0\rangle` isn't part of this subset since the first component isn't
`1\!`, meaning that this subset of vectors isn't closed under scalar multiplication.

Hence, this subset of vectors isn't a linear subspace.

(c)

The following counterexample demonstrates that the set of vectors
`\langle b_(1),\, b_(2),\, b_(3)\rangle \in \mathbb{R}^(3)` where
`b_(1)\, b_(2)\, b_(3) = 0` isn't closed under vector addition.

Consider vectors
`\mathbf{u} = \langle 1,\, 0,\, 0\rangle` and
`\mathbf{v} = \langle 0,\, 1,\, 1\rangle`. While both vectors satisfy the requirement that the scalar product of the components is
`0`, their sum
`(\mathbf{u} + \mathbf{v}) = \langle 1,\, 1,\, 1\rangle` doesn't. Hence, this subset of vectors isn't closed under vector addition and isn't a linear subspace.