What shapes allow us to commute a product?

Prove non-commutativity by example!

(*AB*)^{T} = ?

onto = fills up entire codomain

1-1 (aka 1-to-1) = "non-scrunching"

linear transformation *R*^{3} → *R*^{4} can't be *onto*: there aren't enough columns to span the codomain.

(Though a nonlinear transformation *could* be: example.)

linear transformation *R*^{4} → *R*^{3} *has* to "scrunch" (can't be 1-to-1).

Def: A transformation *R*^{n} → *R*^{n} such that *T*(*x*) = *x* for all *x*
is called the *identity* transformation (from *R*^{n} → *R*^{n}).

What is its matrix, *I*_{n}? (Where does T send the "*e*_{i}" vectors?)

Suppose A is the matrix of a transformation *R*^{n} → *R*^{m}

Under what circumstances could the transformation be "undone"?

In other words, when could we find a transformation *R*^{m} → *R*^{n} with matrix B such that
BA gets us back to where we started?

What about a transformation with a *wide* matrix?

Nope: scrunching can't be undone.

What about a transformation with a *tall* matrix?

Maybe ...

Could we ever have an undoing matrix in *both* directions?

For a square matrix, if BA = I, then AB = I also.

a matrix B with these properties is called an *inverse* of A, written *A*^{ − 1}

Q: When does a square matrix have an inverse? When it doesn't scrunch.

When does a 2x2 matrix not scrunch?

Determinant of 2x2.

Formula for 2x2 inverse. Demonstrate by brute force.

When A has an inverse, it is called invertible, or non-singular.

Thm 5:

If A is an invertible nxn matrix, then for any b, Ax=b has a unique solution, and it's *A*^{ − 1}*b*.

Thm 6:

a.
If A invertible, *A*^{ − 1} is invertible too and (*A*^{ − 1})^{ − 1} = *A*

b. If A and B are invertible nxn matrices, then AB ....

c.
If A is invertible, so is *A*^{T} and (*A*^{T})^{ − 1} = ...

Stacking "Ax=B"s

Augment A with identity and RRE ...

Discuss.