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 plane within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 1.

Are all of the vectors in \(S\) also in \(\IR^3\text{?}\)

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 plane within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 2.

Let \(\vec{z}\) be the additive identity in \(\IR^3\text{.}\) Is \(\vec{z} \in S\text{?}\)

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 plane within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 3.

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

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 plane within \(\IR^3\text{.}\) Call this plane \(S\text{.}\)

Part 4.

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

A subset of a vector space is called a subspace if it is a vector space on its own. The operations of addition and scalar from the parent vector space are inherited by the subspace.

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.

Any subset \(S\) of a vector space \(V\) that contains the additive identity \(\vec 0\) satisfies the eight vector space properties in Definition 2.1.3 automatically, since the operations were well-defined for the parent vector space.

However, to verify that it's a subspace, we still need to check that addition and multiplication still make sense using when only vectors from \(S\) are allowed to be used. So we need to check two things:

• The set 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 set 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 \(S\text{.}\)

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

Part 1.

Let \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\) and \(\vec{w} = \left[\begin{array}{c} a \\ b \\ c \end{array}\right] \) be vectors in \(S\text{,}\) so \(x+2y+z=0\) and \(a+2b+c=0\text{.}\) Show that \(\vec v+\vec w = \left[\begin{array}{c} x+a \\ y+b \\ z+c \end{array}\right]\) also belongs to \(S\) by verifying that \((x+a)+2(y+b)+(z+c)=0\text{.}\)

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

Part 2.

Let \(\vec{v}=\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\in S\text{,}\) so \(x+2y+z=0\text{.}\) Show that \(c\vec v=\left[\begin{array}{c}cx\\cy\\cz\end{array}\right]\) also belongs to \(S\) for any \(c\in\IR\) by verifying an appropriate equation.

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

Part 3.

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

Let \(S=\setBuilder{\left[\begin{array}{c} x \\ y \\ z \end{array}\right]}{ x+2y+z=4}\text{.}\) Choose a vector \(\vec v=\left[\begin{array}{c} \unknown\\\unknown\\\unknown \end{array}\right]\) in \(S\) and a real number \(c=\unknown\text{,}\) and show that \(c\vec v\) isn't in \(S\text{.}\) Is \(S\) a subspace of \(\IR^3\text{?}\)

Since \(0\) is a scalar and \(0\vec{v}=\vec{z}\) for any vector \(\vec{v}\text{,}\) a nonempty set that is closed under scalar multiplication must contain the zero vector \(\vec{z}\) for that vector space.

Put another way, you can check any of the following to show that a nonempty subset \(W\) isn't a subspace:

• Show that \(\vec 0\not\in W\text{.}\)

• 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 \(c\vec v\not\in W\text{.}\)

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

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

• 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 these subsets of \(M_{2 \times 2}\text{,}\) the vector space of all \(2 \times 2\) matrices with real entries. Show that each of these sets is or is not a subspace of \(M_{2 \times 2}\text{.}\)

Part 1.

Consider these subsets of \(M_{2 \times 2}\text{,}\) the vector space of all \(2 \times 2\) matrices with real entries. Show that each of these sets is or is not a subspace of \(M_{2 \times 2}\text{.}\)

Part 2.

Consider these subsets of \(M_{2 \times 2}\text{,}\) the vector space of all \(2 \times 2\) matrices with real entries. Show that each of these sets is or is not a subspace of \(M_{2 \times 2}\text{.}\)

Part 3.

Consider these subsets of \(M_{2 \times 2}\text{,}\) the vector space of all \(2 \times 2\) matrices with real entries. Show that each of these sets is or is not a subspace of \(M_{2 \times 2}\text{.}\)

Part 4.

Let \(W\) be a subspace of a vector space \(V\text{.}\) How are \(\vspan W\) and \(W\) related?

1. \(\vspan W\) may include vectors that aren't in \(W\)

2. \(W\) may include vectors that aren't in \(\vspan W\)

3. \(W\) and \(\vspan W\) always contain the same vectors

If \(S\) is any subset of a vector space \(V\text{,}\) then since \(\vspan S\) collects all possible linear combinations, \(\vspan S\) is automatically a subspace of \(V\text{.}\)

In fact, \(\vspan S\) is always the smallest subspace of \(V\) that contains all the vectors in \(S\text{.}\)