A subset \(S\) of a vector space is called a subspace provided it is equal to the span of a set of vectors from that space.

Consider two non-colinear vectors in \(\IR^3\text{.}\) If we look at all linear combinations of those two vectors (that is, their span), we end up with a planar subspace within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 1.

For any unspecified \(\vec{u}, \vec{v} \in S\text{,}\) is it the case that \(\vec{u} + \vec{v} \in S\text{?}\)

1. Yes.

2. No.

Consider two non-colinear vectors in \(\IR^3\text{.}\) If we look at all linear combinations of those two vectors (that is, their span), we end up with a planar subspace within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 2.

For any unspecified \(\vec{u} \in S\) and \(c\in\IR\text{,}\) is it the case that \(\vec{u}+\left[\begin{array}{c} c \\ c \\ c \end{array}\right] \in S\text{?}\)

1. Yes.

2. No.

Consider two non-colinear vectors in \(\IR^3\text{.}\) If we look at all linear combinations of those two vectors (that is, their span), we end up with a planar subspace within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 3.

For any unspecified \(\vec{u} \in S\) and \(c\in\IR\text{,}\) is it the case that \(c\vec{u} \in S\text{?}\)

1. Yes.

2. No.

A subset \(S\) of a vector space is a subspace provided:

• the subset is closed under addition: for any \(\vec{u},\vec{v} \in S\text{,}\) the sum \(\vec{u}+\vec{v}\) is also in \(S\text{.}\)

• the subset is closed under scalar multiplication: for any \(\vec{u} \in S\) and scalar \(c \in \IR\text{,}\) the product \(c\vec{u}\) is also in \(S\text{.}\)

Note the similarities between a planar subspace spanned by two non-colinear vectors in \(\IR^3\text{,}\) and the Euclidean plane \(\IR^2\text{.}\) While they are not the same thing (and shouldn’t be referred to interchangably), algebraists call such similar spaces isomorphic; we’ll learn what this means more carefully in a later chapter.

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 1.

Let’s assume that \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) and \(\vec{w} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] \) are in \(S\text{.}\) What are we allowed to assume?

1. \(x+2y+z=0\text{.}\)

2. \(a+2b+c=0\text{.}\)

3. Both of these.

4. Neither of these.

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 2.

Which equation must be verified to show that \(\vec v+\vec w = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\) also belongs to \(S\text{?}\)

1. \((x+a)+2(y+b)+(z+c)=0\text{.}\)

2. \(x+a+2y+b+z+c=0\text{.}\)

3. \(x+2y+z=a+2b+c\text{.}\)

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 3.

Use the assumptions from (a) to verify the equation from (b).

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 4.

Is \(S\) is a subspace of \(\IR^3\text{?}\)

1. Yes

2. No

3. Not enough information

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 5.

Show that \(k\vec v=\left[\begin{array}{c}kx\\ky\\kz\end{array}\right]\) also belongs to \(S\) for any \(k\in\IR\) by verifying \((kx)+2(ky)+(kz)=0\) under these assumptions.

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=0}\text{.}\)

Part 6.

Is \(S\) is a subspace of \(\IR^3\text{?}\)

1. Yes

2. No

3. Not enough information

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\)

Part 1.

Which of these statements is valid?

1. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in S\text{,}\) so \(S\) is a subspace.

2. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\in S\text{,}\) so \(S\) is not a subspace.

3. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in S\text{,}\) so \(S\) is a subspace.

4. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 2 \\ 2 \\ 2 \end{array}\right]\not\in S\text{,}\) so \(S\) is not a subspace.

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\)

Part 2.

Which of these statements is valid?

1. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in S\text{,}\) so \(S\) is a subspace.

2. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) and \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\in S\text{,}\) so \(S\) is not a subspace.

3. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in S\text{,}\) so \(S\) is a subspace.

4. \(\left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right]\in S\text{,}\) but \(\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]\not\in S\text{,}\) so \(S\) is not a subspace.

In summary, you can check any of the following to show that a nonempty subset \(W\) isn’t a subspace:

• Find \(\vec u,\vec v\in W\) such that \(\vec u+\vec v\not\in W\text{.}\)

• Find \(c\in\IR,\vec v\in W\) such that \(c\vec v\not\in W\text{.}\)

• Show that \(\vec 0\not\in W\) (same as the last step, with \(c=0\)).

If you cannot do any of these, then \(W\) can be proven to be a subspace by doing both of the following:

1. Prove that \(\vec u+\vec v\in W\) whenever \(\vec u,\vec v\in W\text{.}\)

2. Prove that \(c\vec v\in W\) whenever \(c\in\IR,\vec v\in W\text{.}\)

Consider these subsets of \(\IR^3\text{:}\)

Part 1.

Show \(R\) isn’t a subspace by showing that \(\vec 0\not\in R\text{.}\)

Consider these subsets of \(\IR^3\text{:}\)

Part 2.

Show \(S\) isn’t a subspace by finding two vectors \(\vec u,\vec v\in S\) such that \(\vec u+\vec v\not\in S\text{.}\)

Consider these subsets of \(\IR^3\text{:}\)

Part 3.

Show \(T\) isn’t a subspace by finding a vector \(\vec v\in T\) such that \(2\vec v\not\in T\text{.}\)

Consider the following two sets of Euclidean vectors:

Explain why one of these sets is a subspace of \(\mathbb{R}^2\) and one is not.

Consider the following attempted proof that

Let \(\left[\begin{array}{c} x \\ y \end{array}\right]\in U\text{,}\) so we know that \(x+y=xy\text{.}\) We want to show \(k\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{c} kx \\ ky \end{array}\right]\in U\text{,}\) that is, \((kx)+(ky)=(kx)(ky)\text{.}\) This is verified by the following calculation:

\begin{align*} (kx)+(ky)&=(kx)(ky)\\ k(x+y)&=k^2xy\\ 0[k(x+y)]&=0[k^2xy]\\ 0&=0 \end{align*}

Is this reasoning valid?

1. Yes

2. No

Proofs of an equality \(\mathrm{LEFT}=\mathrm{RIGHT}\) should generally be of one of these forms:

1. Using a chain of equalities:

   \begin{align*} \mathrm{LEFT} &= \cdots\\ &= \cdots\\ &= \cdots\\ &= \mathrm{RIGHT} \end{align*}
   Alternatively:
   \begin{align*} \mathrm{LEFT} &= \cdots & \mathrm{RIGHT} &=\cdots\\ &= \cdots & &= \cdots\\ &= \cdots & &= \cdots\\ &= \mathrm{SAME}& &= \mathrm{SAME} \end{align*}

2. When the assumption \(\mathrm{THIS}=\mathrm{THAT}\) is already known or assumed to be true :

   \begin{align*} && \mathrm{THIS} &= \mathrm{THAT}\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \cdots &= \cdots\\ & \Rightarrow& \mathrm{LEFT} &= \mathrm{RIGHT} \end{align*}