HW Review
Problems 1a, and 2 from the homework gave us a look at two
interesting things we can do with vectors:
- We can form a linear combination of (column) vectors.
-
We can define a function that map vectors to vectors, where
the components of the output vector are fixed linear
combinations of the input vector.
Note: these functions are called linear
transformations
(but more on that later).
Problem 1b gave us an interesting kind of problem: find
coefficients $x_1$, $x_2$, $x_3$ that make the following equality hold:
$$
x_1\cdot
\begin{bmatrix}
-1\\
0\\
2
\end{bmatrix}
+
x_2\cdot
\begin{bmatrix}
0\\
3\\
3
\end{bmatrix}
+
x_3\cdot
\begin{bmatrix}
6\\
3\\
-9
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
$$
So we have a "linear combination equation". We could also have
written down "linear transformation equations", like:
$$
f(\boldsymbol{x}) =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix}
, \text{ for function }
f(\boldsymbol{x}) =
\begin{bmatrix}
\boldsymbol{w_1}\cdot \boldsymbol{x}\\
\boldsymbol{w_2}\cdot \boldsymbol{x}\\
\boldsymbol{w_3}\cdot \boldsymbol{x}
\end{bmatrix}
, \text{ where }
\begin{array}{l}
\boldsymbol{w_1} = [-1\ 0\ \ \ 6\ ]\\
\boldsymbol{w_2} = [\ \ \ 0\ 3\ \ \ 3\ ]\\
\boldsymbol{w_3} = [\ \ \ 2\ 3\ -9]
\end{array}.
$$
ACTIVITY
We can expand for the "linear combination equation" and the
"linear transformation equation" into a system of regular old
high school algebra equations. Write down those systems!
In both cases, you should have gotten the exact same thing:
$$
\begin{array}{rcrcrcl}
-1\cdot x_1 &+&0\cdot x_2 &+& 6\cdot x_3&=0\\
0\cdot x_1 &+&3\cdot x_2 &+& 3\cdot x_3&=0\\
2\cdot x_1 &+&3\cdot x_2 &+& -9\cdot x_3&=0
\end{array}
$$
This kind of object is a
system of linear equations.
Introducing ... the Matrix
So now we have three mathematical objects,
- linear combinations of vectors
- linear transformations (special kinds of functions on
vectors), and
- systems of linear equations
and if you look at our examples, you will see that all three
revolve around the same two-dimensional table of coefficients:
$$
\begin{bmatrix}
-1&0&6\\
0&3&3\\
2&3&-9
\end{bmatrix}
$$
This kind of object is called a
matrix
(plural,
matrices) and by studying matrices we are
simultanesly studying linear combinations, linear
transformations, systems of linear equations, and much more. It
is hard to overstate the ubiquity of matrices in mathematics,
science and engineering.
Matrix $M$ of dimension $m\times n$ (meaning
$m$ rows and $n$ columns) over ring $R$ is the sequence of
values $m_{1,1},\ldots m_{1,n},m_{2,1},\ldots, m_{m,n}$ from $R$.
We typically write matrix $M$ as
$$
\begin{bmatrix}
m_{1,1} & m_{1,2} & \cdots & m_{1,n}\\
m_{2,1} & m_{2,2} & \cdots & m_{2,n}\\
\vdots & \vdots & \ddots & \vdots\\
m_{m,n} & m_{m,2} & \cdots & m_{m,n}
\end{bmatrix}
$$
By the row vectors of $M$ we mean the $m$
vectors $\boldsymbol{v_1},\ldots,\boldsymbol{v_m}$ where
$\boldsymbol{v_i} =
\begin{bmatrix}m_{i,1}& m_{i,2}&\ldots& m_{i,n}\end{bmatrix}$.
By the column vectors of $M$ we mean the $n$
vectors $\boldsymbol{w_1},\ldots,\boldsymbol{w_m}$ where
$\boldsymbol{w_j} =
\begin{bmatrix}
m_{1,j}\\
m_{2,j}\\
\vdots\\
m_{m,j}
\end{bmatrix}$.
ACTIVITY
- Consider the matrix
$$
M_1 =
\begin{bmatrix}
3 & 0 & 2 & 5\\
-1 & 3 & -2 & 3\\
4 & 9 & 0 & -5\\
\end{bmatrix}
$$
- What is the dimension of this matrix?
- What are the row vectors of this matrix?
- What are the column vectors of his matrix?
-
Suppose we want to define the meaning of
"matrix-vector product" $M\cdot \boldsymbol{x}$, for
matrix $M$ and column vector
$\boldsymbol{x}$,
so that it is equivalent to
- the linear combination of the
column vectors of $M$ where the coefficients are the
components of $\boldsymbol{x}$, or
- the linear transformation defined by row vectors of
$M$, applied to the vector $\boldsymbol{x}$.
... hopefully you realize that both of these are in fact
equivalent.
-
Looking at $M_1$ in the above example,
what dimension must the column vector $\boldsymbol{x}$
have in order to have the linear combination of the
columns (or dot products defining the linear
transformation) make sense?
-
In general, if $M$ is an $m\times n$ matrix, in order
for the product $M \cdot \boldsymbol{x}$
make sense, what dimension must the column vector $\boldsymbol{x}$
have?
-
Assuming $M$ and $\boldsymbol{x}$ have compatible
dimensions, how should result of the matrix vector product
$M \cdot \boldsymbol{x}$ be defined?
Matrix vector product
If $M$ is an $m\times n$ matrix
with row vectors $\boldsymbol{r_1},\ldots,\boldsymbol{r_m}$,
and $\boldsymbol{x}$ is column
vector of dimension $n$, the matrix-vector product
$M \cdot \boldsymbol{x}$ is the column vector of dimension $m$
whose $i$th component is $\boldsymbol{r_i}\cdot
\boldsymbol{x}$.
In other words:
$$
M\cdot\boldsymbol{x} =
\left[
\begin{array}{c}
\boldsymbol{r_1}\cdot\boldsymbol{x}\\
\boldsymbol{r_2}\cdot\boldsymbol{x}\\
\vdots\\
\boldsymbol{r_m}\cdot\boldsymbol{x}\\
\end{array}
\right]
$$
Just remember: "row times column".
$$
\begin{bmatrix}
3 & 0 & 2 & 5\\
-1 & 3 & -2 & 3\\
4 & 9 & 0 & -5\\
\end{bmatrix}
\cdot
\begin{bmatrix}
1\\
-1\\
2\\
0
\end{bmatrix}
=
\begin{bmatrix}
\begin{bmatrix}3 & 0 & 2 & 5\end{bmatrix}\cdot\begin{bmatrix}1 & -1 & 2 & 0\end{bmatrix}\\
\begin{bmatrix}-1 & 3 & -2 & 3\end{bmatrix}\cdot\begin{bmatrix}1 & -1 & 2 & 0\end{bmatrix}\\
\begin{bmatrix}4 & 9 & 0 & -5\end{bmatrix}\cdot\begin{bmatrix}1 & -1 & 2 & 0\end{bmatrix}\\
\end{bmatrix}
=
\begin{bmatrix}
7\\
-8\\
-5
\end{bmatrix}
$$
ACTIVITY
-
Compute the following matrix products:
$$
\begin{array}{ccccc}
\begin{bmatrix}
4&0&-2\\
1&-3&4
\end{bmatrix}
\cdot
\begin{bmatrix}
2\\
-1\\
1/2
\end{bmatrix}
= \ \ \ \ \ &,&
\begin{bmatrix}
2&3\\
-1&0\\
5&2
\end{bmatrix}
\cdot
\begin{bmatrix}
2\\
1/2
\end{bmatrix}
= \ \ \ \ \ &,&
\begin{bmatrix}
3&-2\\
5&4
\end{bmatrix}
\cdot
\begin{bmatrix}
-7\\
4
\end{bmatrix}
\end{array}
=
$$
-
Do the following two multiplications:
$$
\begin{array}{cccc}
%%%%%
\begin{bmatrix}
3&-4\\
2&5\\
1&-1
\end{bmatrix}
\cdot
\begin{bmatrix}
1\\
0
\end{bmatrix}
= \ \ \ \ \ &,&
%%%%%
\begin{bmatrix}
3&-4\\
2&5\\
1&-1
\end{bmatrix}
\cdot
\begin{bmatrix}
0\\
1
\end{bmatrix}
=
%%%%%
\end{array}
$$
-
Make a hypothesis: Let $M$ by an $m\times n$ matrix and
$\boldsymbol{x}$ be an $n$-dimensional column vector in
which the $i$th component is $1$, and all other components
are $0$. Then $M\cdot \boldsymbol{x} = $ ....
-
Find a $2\times 2$ matrix with the property that when you
multiply it and any 2-dimensional vector $\boldsymbol{x}$,
you get that same $\boldsymbol{x}$ back again. I.e.
$$
\begin{bmatrix}
?&?\\
?&?
\end{bmatrix}
\cdot
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
=
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}
$$
The unit vectors
In vector space $\mathbb{R}^n$ we define the
unit vectors
$\boldsymbol{e_1},\ldots,\boldsymbol{e_n}$ as
follows: the $j$th component of $\boldsymbol{e_i}$ is 1 if $j=i$
and $0$ otherwise.
From part 3 of the activity, we get the following:
for any $m\times n$ matrix $M$, $M\cdot \boldsymbol{e_i} = $
the $i$th column vector of $M$.
The identity matrix
Let's follow on from part 4 of the activity.
Consider the equation:
$$
M\cdot\boldsymbol{x}=\boldsymbol{x}
$$
If $\boldsymbol{x}$ is an $n$-dimensional vector, $M$ must have
dimension $n\times n$ (this means it is
square), since
it has to have $n$ columns for multiplication by
$\boldsymbol{x}$ to make sense, and it has to have $n$ rows in
order to produce an output vector of dimension $n$.
We are looking for the $n\times n$
identitymatrix,
$I$,
which should have the property that the above equation holds for
all $\boldsymbol{x}$. This means, in particular, that
$$
I\cdot\boldsymbol{e_i}=\boldsymbol{e_i}.
$$
But from the previous section, $I\cdot\boldsymbol{e_i}=$ the
$i$th column of $I$. So, the $i$th column of $I$ is
$\boldsymbol{e_i}$.
This means that the $i$th column of $I$ is 1 at the the $i$th
row and zero everywhere else.
So $I$ has the shape we hypothesized in class: 1's along the
main diagonal, and zeros everywhere else. For example, the
$3\times 3$ identity matrix is :
$$
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}
$$
Christopher W Brown