Lecture 1 - Intro to The Course

Syllabus, syllabus, syllabus for anything related to the course. We have HW due next week, so you can take it easy and just read. It'll be posted tonight.

As a prerequisite (for our online viewers), go checkout Lecture 1 - Syllabus and Whatnot and go through those lecture notes for more on this. The document Final Review Sheet of Theorems does a good overview of important theorems that were hard to remember last quarter.

We skipped some things from 3.E, and other sections. We're going to focus on those sections to help recover some things lost from MATH 306. That'll teach some new material, while focusing on reviewing things from that course.

(Review) Getting Started: What's $V$ ?

Recall what the definition of a vector space is:

Vector Space

A vector space is a set $V$ along with an addition on $V$ and a scalar multiplication on $V$ such that the following properties hold:

Commutativity: $u + v = v + u$ for all $u, v \in V$ .
Associativity: $(u + v) + w = u + (v + w)$ and $(a b) v = a (b v)$ for all $u, v, w \in V$ and all $a, b \in F$ .
Additive Identity: There exists an element $0 \in V$ such that $v + 0 = v$ for all $v \in V$ .
Additive Inverse: For every $v \in V$ there exists $w \in V$ such that $v + w = 0$ .
Multiplicative Identity: $1 v = v$ for all $v \in V$ .
Distributive Properties: $a (u + v) = a u + a v$ and $(a + b) v = a v + b v$ for all $a, b \in F$ and all $u, v \in V$ .

and a subspace:

Subspace

A subset $U \subseteq V$ is called a subspace of $V$ if $U$ is also a vector space (using the same addition and scalar multiplication as on $V$ ).

Where we do the following to verify them:

Conditions for a subspace

A subset $U \subseteq V$ is a subspace of $V$ iff $U$ satisfies the following three conditions:

Additive Identity: $\vec{0} \in U$
Closed Under Addition: $\vec{u}, \vec{w} \in U$ implies that $\vec{u} + \vec{w} \in U$
Closed Under Scalar Multiplication: $a \in F$ and $\vec{u} \in U$ implies that $a \vec{u} \in U$

Consider the Following...

Say $V$ is a vector space and $U_{1}, U_{2}$ are subspaces. What is $U_{1} + U_{2}$ ? We defined it as:

Sum of Subsets

Suppose $U_{1}, . . ., U_{m}$ are subsets of $V$ . The sum of $U_{1}, . . ., U_{m}$ , denoted $U_{1} + . . . + U_{m}$ , is the set of all possible sums of elements of $U_{1}, . . ., U_{m}$ . More precisely:
$U_{1} + \dots + U_{m} = {u_{1} + \dots + u_{m} : u_{1} \in U_{1}, . . ., u_{m} \in U_{m}}$

It's essentially the smallest subspace containing both $U_{1}$ and $U_{2}$ .

Not a Union

Why not just say $U_{1} + U_{2} = U_{1} \cup U_{2}$ ? Consider the following counter example. If $U_{1}, U_{2}$ are different lines through the origin in $R^{2}$ . The union would contain both lines, but then you could add vectors to get a point off the line.

Direct Sum

When is $U_{1} + U_{2}$ a direct sum? It's a direct sum if there's a unique way to write the vector, composed of $u_{1}, u_{2}$ :

Direct Sum

Suppose $U_{1}, . . ., U_{m}$ are subspaces of $V$ .

The sum $U_{1} + \dots + U_{m}$ is called a direct sum if each element of $U_{1} + \dots + U_{m}$ can be written in only one way as a sum $u_{1} + \dots + u_{m}$ , where each $u_{j} \in U_{j}$ .
If $U_{1} + \dots + U_{m}$ is a direct sum, then $U_{1} \oplus \dots \oplus U_{m}$ denotes $U_{1} + \dots + U_{m}$ where the $\oplus$ notation serves as an indication that this is a direct sum.

Consider the blue vector in the above drawing. It requires the unique blue vectors from $U_{1}, U_{2}$ correspondingly. We can just show that if the zero vector $\vec{0}$ can be written uniquely, then it's equivalent to the above condition. This is because our vector:

v = u_{1} + u_{2} \Rightarrow \vec{0} = u_{1} + u_{2} - v

Which is unique. The theorem:

Condition for a direct sum

Suppose $U_{1}, \dots, U_{m}$ are subspaces of $V$ . Then $U_{1} + \dots + U_{m}$ is a direct sum iff the only way to write $0$ as a sum $u_{1} + \dots + u_{m}$ , where each $u_{j} \in U_{j}$ . is by taking each $u_{j} = 0$ .

Summarizes this, and has a corresponding proof.

An easy condition to determine this is the following:

Direct sum of two subspaces

Suppose $U$ and $W$ are subspaces of $V$ . Then $U + W$ is a direct sum iff $U \cap W = {0}$ .

(NEW) Products of Vector Spaces and Quotients

Question: Given subspaces $U_{1}, . . ., U_{m}$ of a vector space $V$ , how do we construct of $U_{1} + U_{2} + \dots + U_{m}$ ? We talked about it; we just take a vector from each space and add them up:

v = u_{1} + \dots + u_{m}, u_{i} \in U_{i}

for all $i \in N^{\leq m}$ .

But notice we have an $m$ tuple of choices for our $u_{i}$ 's! That's similar to a Cartesian Product. Notice our choice can be represented by:

(u_{1}, . . ., u_{m}) \in U_{1} \times \dots \times U_{m}

Thus, we just choose an element of $U_{1} \times \dots \times U_{m}$ to make this construction. We know this is a set, but let's talk about it as a vector space. Is it one? We first need to define this operation, but it's pretty clear what it should be. You add component wise, and distribute the scalar through all components, just like a vector in $R^{m}$ :

Product Space

Suppose $V_{1}, . . ., V_{m}$ are all vector spaces over the same field $F$ . The product:

V_{1} \times \dots \times V_{m} = {(v_{1}, . . ., v_{m}) : \forall i (v_{i} \in V_{i})}

is a vector space, where vector addition and scalar multiplication are componentwise, meaning:

(v_{1}, \dots, v_{m}) + (w_{1}, \dots, w_{m}) = (v_{1} + w_{1}, \dots, v_{m} + w_{m})

and:

α (v_{1}, \dots, v_{m}) = (α v_{1}, \dots, α v_{m})

Notice that each of the $+$ signs are defined internally for each $U_{i}$ , possibly separately. We could have $U_{1} = R^{2}$ , $U_{2} = P_{3} (R)$ , and so on.

The proof for this is pretty easy to show, it just takes time to show. Hence, we don't show it here.

Notice the zero vector here is just the tuple of choosing all the zeroes in each $U_{i}$ :

\vec{0} = ({\vec{0}}_{U_{1}}, \dots, {\vec{0}}_{U_{m}})

Using a product space

Let $V = F^{2, 2} \times P_{2} (F)$ . Recall the former is the set of $2 \times 2$ matrices, and the second is the set of all polynomials with degree $2$ or less.

For instance, one vector is:

v = ([\begin{matrix} - 2 & 7 \\ 5 & 1 \end{matrix}], - 2 + 5 x + x^{2}) \in V

A good question is what's a valid basis vector (let's just say standard basis) for $V$ ? One example is:

v_{1} = ([\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], 0)

We don't use basis between them (although technically you can). A basis for this space would be:

β = {([\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}], 0), ([\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}], 0), \dots, ([\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}], 1), ([\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}], x), ([\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}], x^{2})}

It's interesting there's 7 vectors here, and that's the sum of the dimension of the cartesian product subspaces.

(Review) Getting Started: What's V?

Consider the Following...

Direct Sum

(NEW) Products of Vector Spaces and Quotients

(Review) Getting Started: What's $V$ ?