Lecture 14 - Linear Transformations ++

Recall that $T : V \to W$ is linear iff:

$T (v + w) = T v + T w$ , so the linear transformation is additive (has additivity)
$T (α v) = α T v$ , so the linear transformation is homogeneous (has homogeneity)

Recall that the zero map $Z$ maps all vectors to $\vec{0}$ , so $Z (v) = \vec{0}$ . But the interesting thing is to consider some set of vectors that all get mapped to ${\vec{0}}_{W}$ :

Nullspace, Range

Given $T \in L (V, W)$ , define:

$null (T) = {v \in V : T (v) = \vec{0}}$
$range (T) = {w \in W : \exists v \in V s . t . T (v) = w} = {T (v) : v \in V}$
Which are the nullspace or kernel of $T$ , and the range/image of $T$ respectively.

As an example, in $R^{2}$ let $T$ be the orthogonal projection onto the line $y = m x$ :

Then:

The $range (T)$ is any vector on the line $y = m x$ . The line contains the vector $(1, m)$ (from rise over run) so then $range (T) = span (⟨ 1, m ⟩)$ .
The $null (T)$ is the vectors perpendicular to the line $y = m x$ . The line perpendicular is $y = \frac{- 1}{m} x$ , so then we can choose the vector spanned via $null (T) = span (⟨ m, - 1 ⟩)$ .

But notice that both of these were their own subspaces!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Always, you can find out that $null (T)$ and $range (T)$ are nullspaces of $V$ . We'll see later that they are a direct sum even iff $T : V \to V$ (and not $W$ ). We'll prove this via projections which we'll see in two weeks.

Recall that $dim (R^{2}) = 2$ and $dim (range (T)) + dim (null (T))$ since each dimension solely was 1. But does this happen all the time? Yes!!! This is the rank-nullity theorem (you've heard of this) and we'll prove it:

The Fundamental Theorem of Linear Maps

Given $T \in L (V, W)$ where $V$ is finite dimensional, then:

dim (V) = dim (range (T)) + dim (null (T))

We won't do the whole proof here, but here's the idea.

Proof

Suppose that $dim (V) = n$ .

1. Start with a basis for $\text{null}(T)$, denoted $v_1, ..., v_k$. 2. Extend the basis to encompass all of $V$ via $v_1, ..., v_k, v_{k+1}, ..., v_n$ 3. Find the basis vectors $T(v_{k+1}), ..., T(v_n)$ as a basis for $\text{range}(T)$, and show that it is a basis. 1. This is where the heavy lifting is. 2. Show LI. Consider: $$ a_1T(v_{k+1}) + \dots + a_{n-k}T(v_{n}) = \vec{0} $$ But then we can rewrite our transformation as: $$ T(a_1v_{k+1} + \dots + a_{n-k}v_{n}) = \vec{0} $$ So then $a_1v_{k+1} + \dots + a_{n-k}v_n \in \text{null}(T)$. But we have a basis for it so then there are scalars such that $v = a_1v_{k+1} + \dots + a_{n-k}v_n = b_1v_1 + \dots + b_kv_k$ for some $b_i$'s. So then: $$ \vec{0} = b_1v_1 + \dots + b_kv_k - a_1v_{k+1} - \dots - a_{n-k}v_k $$ But this $v_1, ..., v_k, ..., v_n$ is a basis for $V$, so then all $a_i, b_i = 0$, so then $v = \vec{0}$ and thus we get LI. 1. Show they span (this is omitted for times sake). 4. Thus then $\text{dim}(\text{range}(T)) = n-k$, as needed. ☐