Lecture 4 - (Online) Subspaces and Constructions

Subspaces

Let's just start with the definition:

Subspace

A subspace $U$ of a vector space $V$ if $U$ is also a vector space (under the same operations of $U$ )

The question here is what properties of a subspace are guaranteed given that $V$ itself is a vector space? We expect many properties to carry over to $U$ . There are some things to consider:

The $\vec{0}$ may not be in $U$ . Being a subspace doesn't guarantee it.
We need to be careful that $\vec{u} + \vec{v} \in U$ for any $\vec{u}, \vec{v} \in U$ . It's possible that we leave $U$ and enter the uncovered part of $U$ in $V$ .
Similarly, we may "leave" $U$ when doing scalar multiplication of $\vec{u}$ .

Theorem

A subset $U$ of a vector space $V$ iff:

Additive Identity: $\vec{0} \in U$ .
Closed under vector addition: $\forall x, y \in U, x + y \in U$
Closed under scalar multiplication: $\forall x \in U, \forall α \in F, α x \in U$ .

Before we prove this, let's go through some examples:

Example

Let $W = {(x_{1}, x_{2}, 0) : x_{1}, x_{2} \in R}$ . Clearly $W \subseteq R^{3}$ . We claim that this subset is a subspace of $R^{3}$ .

\begin{proof}

Notice $\vec{0} = (0, 0, 0) \in U$ .
Let $\vec{x}, \vec{y} \in U$ be arbitrary, so then we can say that $\vec{x} = (x_{1}, x_{2}, 0), \vec{y} = (y_{1}, y_{2}, 0)$ , for $x_{1}, x_{2}, y_{1}, y_{2} \in R$ . Notice that:

\vec{x} + \vec{y} = (x_{1} + y_{1}, x_{2} + y_{2}, 0) \in U

Let $\vec{x} \in U$ be arbitrary, and likewise for $α \in R$ . We define $\vec{x}$ the same as in (2). Notice that:

α \vec{x} = (α x_{1}, α x_{2}, 0) \in U

Therefore, $U$ is a subspace of $R^{3}$ .
\end{proof}

Example

Recall: $M_{n \times n} (F)$ is the vector space of $n \times n$ matrices, with entries from $F$ . Recall that a matrix is symmetric if it is equal to it's transpose ( $A = A^{T}$ ); the entries are symmetrical about the diagonal. For example:

[\begin{matrix} 0 & 1 & 2 \\ 1 & - 1 & 5 \\ 2 & 5 & 7 \end{matrix}]

We claim that the set of $n \times n$ symmetric matrices are a subspace of $M_{n \times n} F$ .

\begin{proof}
Let's just prove it for $n = 3$ , as the process is largely the same for any $n$ .

Notice $\vec{0}$ is the zero matrix in our subspace:

[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \in S

Let $A, B \in S$ . Consider $A + B \in S$ :

A = [\begin{matrix} a & b & c \\ b & d & e \\ c & e & f \end{matrix}], B = [\begin{matrix} g & h & i \\ h & j & k \\ i & k & l \end{matrix}], A + B = [\begin{matrix} a + g & b + h & c + i \\ b + h & d + j & e + k \\ c + i & e + k & f + l \end{matrix}] \in S

Let $α \in F$ . Then:

α A = [\begin{matrix} α a & α b & α c \\ α b & α d & α e \\ α c & α e & α f \end{matrix}] \in S

\end{proof}

R^{R}

This is the vector space of functions from $R \to R$ .

\begin{proof}

The zero function is in $R^{R}$ .
... (you can show the rest pretty easily).
\end{proof}

Constructions of New Subspaces from Old Ones

Let $V$ be a vector space. Let $U, W$ be subspaces of $V$ . A couple questions we should answer are:

Is $U \cup W$ a subspace of $V$ ?
Is $U \cap W$ a subspace of $V$ ?

Notice that clearly if (1) is true then (2) must be true, but let's take it one step at a time:

1: Is $U \cup W$ a subspace of $V$ ?

Let's look at an example to see what's going on here:

Example

Let $U = {(x_{1}, x_{2}, 0) : x_{1}, x_{2} \in R}$ and $W = {(0, x_{1}, x_{2}) : x_{1}, x_{2} \in R}$ . These both are subspaces of $R^{3}$ , which itself is a vector space.

But notice that $U \cup W$ isn't closed under vector addition. If we take $u \in U$ and $w \in W$ , then:

u + w = (u_{1}, u_{2}, 0) + (0, w_{1}, w_{2}) = (u_{1}, u_{2} + w_{1}, w_{2}) \notin U \cup W

So $U \cup W$ is not closed under vector addition, so $U \cup W$ is not a subspace of $W$ .

2: Is $U \cap W$ a subspace of $V$ ?

It turns out that $U \cap W$ is indeed a vector space. The proof is in HW 1 - Vector Spaces#1.C TODO (read first).

A new question (Smallest Subspace)

What is the smallest subspace of $V$ that contains both $U \cup W$ ? We know that it's possible that $U \cup W$ to not be a subspace, but what are the conditions that make it not so?

Define a set $S = {u + w : u \in U, w \in W}$ . This is essentially saying the set of vectors that are spanned by anything from $u$ and $w$ . You may note that the smallest subspace of $V$ that contains both $U$ and $W$ must also contain $S$ (by definition)

Is $S$ a subspace? Yes! Why? Let's prove it:

\begin{proof}

$\vec{0} \in S, \vec{0} = \vec{0} + \vec{0}$
Let ${\vec{s}}_{1} = {\vec{u}}_{1} + {\vec{w}}_{1}$ , ${\vec{s}}_{2} = {\vec{u}}_{2} + {\vec{w}}_{2}$ . then ${\vec{s}}_{1} + {\vec{s}}_{2} = ({\vec{u}}_{1} + {\vec{w}}_{1}) + ({\vec{u}}_{2} + {\vec{w}}_{2}) = ({\vec{u}}_{1} + {\vec{u}}_{2}) + ({\vec{w}}_{1} + {\vec{w}}_{2})$ . But look! ${\vec{u}}_{1} + {\vec{u}}_{2} \in U$ and likewise ${\vec{w}}_{1} + {\vec{w}}_{2} \in W$ so then since $S$ is defined as it is, by definition ${\vec{s}}_{1} + {\vec{s}}_{2} \in S$ .
Let $α \in F$ be arbitrary, and $\vec{s} \in S$ . Notice that $α \vec{s} = α (\vec{u} + \vec{w}) = α \vec{u} + α \vec{w}$ . The former vector $α \vec{u} \in U$ and $α \vec{w} \in W$ so then by definition then $α \vec{s} \in S$ .
\end{proof}
Note that $S$ is the smallest subspace containing both $\vec{u}, \vec{w}$ . We give $S$ some notation, namely as the sum of two subspaces $U, W$ :

The Sum of Vector Spaces

Suppose $U_{1}, . . ., U_{m}$ are subspaces of vector space $V$ . The sum of $U_{1}, U_{2}, . . ., U_{m}$ , denoted by $U_{1} + U_{2} + . . . + U_{m}$ , is the set of all possible sums of elements from $U_{1}, . . ., U_{m}$ :

U_{1} + . . . + U_{m} = {{\vec{u}}_{1} + . . . + \vec{u_{m}} | {\vec{u}}_{1} \in U_{1}, . . ., {\vec{u}}_{m} \in U_{m}}

And $U_{1} + . . . + U_{m}$ is the smallest subspace containing $U_{1}, . . ., U_{m}$ .

We use the words "smallest" to mean that there exists no strict subset $F$ from $U_{1}, . . ., U_{m}$ such that $F$ is a vector space while containing all of $U_{1}, . . ., U_{m}$ .

Subspaces

Constructions of New Subspaces from Old Ones

1: Is U∪W a subspace of V?

2: Is U∩W a subspace of V?

A new question (Smallest Subspace)

1: Is $U \cup W$ a subspace of $V$ ?

2: Is $U \cap W$ a subspace of $V$ ?