Lecture 18 - Finishing Matrices, Starting Invertibility

From Lecture 17 - Continuing Matrices#^62ce98 we defined matrix multiplication. Hence, we make the following definition:

Matrix Multiplication

Given an $m \times n$ matrix $A$ and a $n \times p$ matrix $B$ , we define $A B$ to be the $m \times p$ matrix whose $i j$ entry is given by:

c_{i j} = \sum_{k = 1}^{n} A_{i k} B_{k j}

Some notation is:

$A_{i, \cdot}$ : the $i$ -th row of $A$
$A_{\cdot, j}$ : the $j$ -th column of $A$
Notice that the dot product form is here, so sometimes that's used to shorthand the notation:

A_{i, \cdot} \cdot B_{\cdot, j}

Another fact is that:

(A B)_{i, \cdot} = A (B_{\cdot, j})

3.D: Invertibility

You've heard of inverse functions before, so let's look at the definition then talk more about it:

Inverse Map

A map $T \in L (V, W)$ is invertible if there exists some $S \in L (W, V)$ such that:

S \circ T = I v, T \circ S = I w

where $I$ is the identity map such that $I v = v$ for any $v \in V$ .
Such a map $S$ is called an inverse of $T$ .

Notice again that $S$ is an inverse. But it is true, one of our first theorems, that inverses $S$ are unique.

Uniqueness of Inverses

Inverses are unique.

Proof
Suppose $T$ is some linear transformation from $V$ to $W$ , and that $S_{1}, S_{2}$ are inverses of $T$ . Then both $S_{1} \circ T = T \circ S_{2} = I$ and $S_{2} \circ T = T \circ S_{2} = I$ .

S_{1} = S_{1} \circ I_{w} = S_{1} \circ (T \circ S_{2}) = (S_{1} \circ T) \circ S_{2} = I_{v} \circ S_{2} = S_{2}

where we use the associativity of transformations in the middle.
☐
We denote the inverse of an invertible map $T$ by $T^{- 1}$ . But when is $T$ invertible?

Theorem

$T \in L (V, W)$ is invertible iff $T$ is injective and surjective (ie: $T$ is bijective).

Note that in calculus we looked at functions like $e^{x}$ and found inverses $\ln (x)$ . But we need surjectivity now since the domains and co-domains may be different. Here, from calculus we needed a change in the domain from $x \in R$ to $x \in (0, \infty)$ .
Proof
We'll keep the details out of here and prove this broadly. This is an iff proof so start with $\to$ . Suppose $T$ is invertible. We first show injectivity, so suppose $T v_{1} = T v_{2}$ for some $v_{1}, v_{2} \in V$ . Then since $T$ is invertible, then $T^{- 1}$ exists, and:

T^{- 1} \circ T v_{1} = I_{v} v_{1} = v_{1}

and likewise $T^{- 1} \circ T v_{2} = v_{2}$ , but since $T v_{1} = T v_{2}$ then $v_{1} = v_{2}$ as requested.

For surjectivity, let $w \in W$ be arbitrary. We need to choose some vector in $V$ where $T v = w$ . I claim that $T^{- 1} w$ is the vector to choose since:

T (T^{- 1} w) = I_{w} w = w

thus $T$ is surjective.

Now we'll prove $\leftarrow$ , so suppose $T$ is bijective. We have to produce a map such that we get the properties of our invertibility. We define a map $S : W \to V$ such that:

S (\vec{w}) = \vec{v}

where since $T$ is surjective, then there exists this vector $\vec{v}$ where $T (S (\vec{w})) = \vec{w}$ . We skip over that $S$ is a linear map and $S \in L (W, V)$ but we skip that detail.

We'll need to show that $S \circ T$ is the identity (as we already can see $T \circ S$ is the identity map as we built it that way). Notice that:

T \circ (S \circ T) \vec{v} = (T \circ S) (T \vec{v}) = I_{w} (T \vec{v}) = T \vec{v}

So since $T$ is injective, the argument for $T$ on both sides are equal, so $(S \circ T) \vec{v} = \vec{v}$ so then $S \circ T = I_{v}$ .
☐