Lecture 9 - Linear Independence and Span

We'll cover linear independence and span today.

Span

Define $span (v_{1}, . . ., v_{k})$ , where the argument is a list of vectors, as the set of linear combinations of those vectors, ie:

span (v_{1}, . . ., v_{k}) = {α_{1} v_{1} + \dots + α_{k} v_{k} | v_{1}, . . ., v_{k} \in V, α_{1}, . . ., α_{k} \in F}

Note that:

This collection of the span of some vectors is a subspace. ( $0 \in span$ , closure under addition and scalar multiplication, see Chapter 2 - Finite-Dimensional Vector Spaces#2.A Span and Linear Independence).
It is also the smallest subspace per Chapter 2 - Finite-Dimensional Vector Spaces#Linear Combinations and Span.
If we were to define some subspaces $V_{i} = span (v_{i})$ (ie: scalar multiples of $v_{i}$ ), then we get that $span (v_{1}, . . ., v_{k}) = V_{1} + V_{2} + \dots + V_{k}$ .
There is a unique way to write any vector in $span (v_{1}, . . ., v_{k})$ precisely when $V_{1} + \dots + V_{k}$ is direct!

There's some language here to deal with span:

Terminology

If we say that $span (v_{1}, . . ., v_{k}) = U$ we say that $v_{1}, . . ., v_{k}$ spans $U$ . Another way to say this is to say that $v_{1}, . . ., v_{k}$ is a spanning list for $U$ .

Let's look at some examples:

Example

Let $V$ be the vector space of $2 \times 2$ matrices with real entries, so $V = M_{2 \times 2} (R)$ . Let $U$ be the subset of $2 \times 2$ symmetric matrices with real entries. So $u \in U$ if it's of the form:

u = [\begin{matrix} a & b \\ b & c \end{matrix}]

Is $U$ a subspace? Yes, due to all properties being considered. But can we find a spanning list $u_{1}, . . ., u_{k} \in U$ ?

Proof
Yes you can, using:

u_{1} = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]

u_{2} = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]

u_{3} = [\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}]

We can prove that $span (u_{1}, u_{2}, u_{3}) = U$ . We show both $\subseteq$ and $\supseteq$ . The $\subseteq$ is free since $U$ itself is a subspace, so it must contain its span.

For the latter, we need to show that we can take any vector $v \in U$ such that $v \in span$ . So then it must be of the form:

v = [\begin{matrix} a & b \\ b & c \end{matrix}] = a u_{1} + b u_{2} + c u_{3}

For $a, b, c \in R$ , so then $v \in span (u_{1}, u_{2}, u_{3})$ .
☐
Note that $u_{1}, u_{2}, u_{3}$ isn't the only spanning list. We could add say:

u_{4} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

and still $u_{1}, . . ., u_{4}$ is a spanning list.

But here the uniqueness of the combinations of $u_{1}, u_{2}, u_{3}$ is interesting to us. But what if we had:

u_{1}^{'} = 0.5 u_{1}, u_{2}^{'} = 7 u_{2}, u_{3}^{'} = π u_{3}

These $u_{1}^{'}, u_{2}^{'}, u_{3}^{'}$ is still a spanning list. In general, if you know that $span (v_{1}, . . ., v_{k})$ , then $span (α_{1} v_{1}, . . ., α_{k} v_{k})$ is the same (provided that $α_{i} \neq 0$ ).

So ideally:

We choose spanning lists that have the fewest number of vectors
We choose vectors such that they are non-scaled (although in the grand scheme of things this is arbitrary).

This brings us to...

Linear Independence

$v_{1}, . . ., v_{k}$ are linearly independent if for all $α_{i} \in F$ we have:

α_{1} v_{1} + \dots + α_{k} v_{k} = \vec{0} \Rightarrow \forall i (α_{i} = 0)

We already know $0 v_{1} + \dots + 0 v_{k} = \vec{0}$ . But this definition implies that this is the only way to write $\vec{0}$ . Why is this the case? See Chapter 2 - Finite-Dimensional Vector Spaces#Linear Independence. If you wish to prove that $v_{1}, . . ., v_{k}$ are linearly independent:

Suppose that $α_{1} v_{1} + \dots + α_{k} v_{k} = \vec{0}$ .
From that, determine that all $α_{i} = 0$ .

If and iff

When a definition is given in math, "if" is equivalent to "iff"!

Linear Dependence

$v_{1}, . . ., v_{k}$ is linearly dependent if $v_{1}, . . ., v_{k}$ is not Linear Independent (LI).

The negation is that there exist some $α_{1}, . . ., α_{k}$ not all 0 such that:

α_{1} v_{1} + \dots + α_{k} v_{k} = \vec{0}

Example

Consider $(1, 3), (1, 1), (4, 6)$ . Are these LI or LD? They are dependent since:

(4, 6) = (1, 3) + 3 (1, 1)

But use the definition:

1 (1, 3) + 3 (1, 1) + (- 1) (4, 6) = (0, 0) = \vec{0}

Notice that this is using the definition, still:

0 v_{1} + 0 v_{2} + 0 v_{3} = \vec{0}

Which isn't interesting and doesn't use the definition.

Note that ${}$ (the empty list) is LI, as the definition is vacuously satisfied. Also the list itself is LI or LD, not the vectors themselves.

Suppose I have a collection of vectors $v_{1}, v_{2}, 0, v_{3}$ . This list is LD since we can put any scalar in front of 0:

0 v_{1} + 0 v_{2} + 1779 \cdot \vec{0} + 0 v_{3} = \vec{0}

So any list containing the zero vector is LD.