Linear transformation
We have talked about "linear transformations" without ever
defining them. Now is the time to correct that! As you'll see,
a linear transformation is a function that takes a vector as
input and produces a vector as output, and which satisfies two
key properties.
At first blush, linear transformations seem to have nothing to
do with matrices, which are after all our current topic.
Vectors ... yes. Matrices ... no. But the amazing thing is
that this abstract definition of linear transformation is in
fact exactly equivalent to being a function that is defined as
a matrix times an input vector.
A function $T$ with inputs from vector space $R^n$ and outputs
in vector space $R^m$ is a linear transformation if and only
if there is an $m\times n$ matrix $A$ over $R$ such that
for any $\boldsymbol{u}$ in $R^n$,
$T(\boldsymbol{u}) = A\boldsymbol{u}$.
We have to prove
"$T$ is a linear transformation" $\Leftrightarrow$ "there is an
$A$ such that $T(\boldsymbol{u}) = A\boldsymbol{u}$".
As we usually do, we will prove equivalence $x\Leftrightarrow
y$ by proving both $x\Rightarrow y$ and $y\Rightarrow x$.
Part 1: Prove that if $T(\boldsymbol{u}) = A\boldsymbol{u}$ then $T$ is
a linear transformation.
ACTIVITY:
Prove that for any $m\times n$ matrix $A$: [Break up into
groups and choose one of the below to do.]
-
$(cA)\boldsymbol{u} = A(c \boldsymbol{u})$ for any
scalar $c$ and vector $\boldsymbol{u}$
-
$A(\boldsymbol{u}+\boldsymbol{v})=A\boldsymbol{u} + A\boldsymbol{v}$
Hint: the easiest way to do this is write down
expressions for element $i,j$ of the output for the
two sides and then show why the two ouput expressions are
the same.
Part 2:
Prove that if $T$ is a linear transformation then there is an
$A$ such that $T(\boldsymbol{u}) = A\boldsymbol{u}$.
The $i$th unit
vector in $R^n$, denoted $\boldsymbol{e_i}$, is the
vector with $i$th component 1, and
all other components 0.
Note that for any vector $\boldsymbol{u}$ in $R^n$ we have
$\boldsymbol{u} = u_1\boldsymbol{e_1} + \cdots + u_n\boldsymbol{e_n}$;
in other words, all vectors can be written as linear
combinations of the unit vectors.
Consider the column vectors
$\boldsymbol{c_1},\ldots,\boldsymbol{c_n}$, where
$\boldsymbol{c_i} = T(\boldsymbol{e_i})$, and let $A$
be the matrix with column vectors
$\boldsymbol{c_1},\ldots,\boldsymbol{c_n}$.
| $A\boldsymbol{u} = u_1\boldsymbol{c_1} + \cdots + u_n\boldsymbol{c_n}$, |
this equivalence of matrix-times-vector and linear
combinations of the matrix's column vectors with
coefficients given by the vector's components.
(see "The many meanings of $A\cdot\boldsymbol{x} = \boldsymbol{0}$",
Class 33)
|
| $A\boldsymbol{u} = u_1T(\boldsymbol{e_1}) + \cdots + u_nT(\boldsymbol{e_n})$, |
this follows from our definition of $\boldsymbol{c_i}$ as $\boldsymbol{c_i} = T(\boldsymbol{e_i})$
above
|
| $A\boldsymbol{u} = T(u_1\boldsymbol{e_1}) + \cdots + T(u_n\boldsymbol{e_n})$, |
this follows by applying property i of the
definition of linear transformation to each term from
the previous line
|
| $A\boldsymbol{u} = T(u_1\boldsymbol{e_1} + \cdots + u_n\boldsymbol{e_n})$, |
This follows from property ii of the
definition of linear transformation. Technically,
there's an inductive proof in there, because we have
to combine $n$ applications of $T$ into a single
application, and property ii only states that
a pair of applications of $T$ can be combined into a
single application. Giving this inductive proof will
be one of your homework problems!
|
| $A\boldsymbol{u} = T(\boldsymbol{u})$ |
|
This is a beautiful result! Whenever we have a matrix $A$, it
defines a linear transformation $T(\boldsymbol{u}) =
A\boldsymbol{u}$. And whenever we have a linear
transformation $T$, there is a matrix $A$ that defines it.
But there's more! Sometimes a proof does more than merely
verifiy that the theorem statement is correct. The
interesting proofs provide insights, and this is just such a
proof. If you look at how this result is proved, you see that
it's telling us that a linear transformation is completely
described by how it maps the unit vectors. So even though
there are infinitely many input vectors a linear
transformation has to map, the transformation can be defined
by a small set of vectors: the images of the unit vectors
under that transformation ... and that is precisely what the
matrix $A$ constructed by the proof is.
Square matrics
Consider the collection of all $n\times n$ matrices, for a given
$n$ and ring $R$. Notice that $AB$ results in another $n\times n$
matrix. So we say this collection of matrices is "closed" under
multiplication - the product stays in the same collection. So
for $n\times n$ matrices we have addition, and we have
multipliction. In fact, you can verify that we have all the
ring axioms (though multiplication is not commutative)! So this
is a ring, and therefore we can "do arithmetic". Part 2 of today's
in-class activity is a chance for
you to see how super-simple matrix arithmetic provides an
elegent and efficient mechansim for computing with geometric
objects, as is ubiquitous in computer graphics, computer aided
design, robotics and much more.