Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: N/A

Properties: N/A
Sufficiencies: N/A

Nonsingular Matrix to Model Gravity

Let's experiment with gravity. Suppose we study a falling object (ball). We will measure the height $h$ of the ball at different times in an experiment.

Suppose we collect the following data.
![[Pasted image 20240524134524.png#invert]]
We have three data points:

\begin{aligned} (t_{1}, h_{1}) & = (3.0, 3) \\ (t_{2}, h_{2}) & = (3.3, 2.559) \\ (t_{3}, h_{3}) & = (3.6, 1.236) \end{aligned}

where $t$ is in seconds and $h$ is in meters. Notice that our data takes the form of a parabolic curve. We can guess that we can use a quadratic polynomial to model our data. The general form of a quadratic polynomial for our function $h (t)$ is

h (t) = a_{0} + a_{1} t + a_{2} t^{2}

To create a model of our data, we want to find the coefficients $a_{0}, a_{1}, a_{2}$ that most closely fit our data such that

h_{i} = h (t_{i}) = a_{0} + a_{1} t_{i} + a_{2} t_{i}^{2}

With this in mind, let's plug in our data points.

\begin{aligned} h (t_{1}) & = a_{0} \cdot 1 + a_{1} \cdot t + a_{2} \cdot t^{2} \\ = a_{0} \cdot 1 + a_{1} \cdot 3.0 + a_{2} \cdot 9 = 3 \\ h (t_{2}) & = a_{0} \cdot 1 + a_{1} \cdot 3.3 + a_{2} \cdot 10.89 = 2.559 \\ h (t_{3}) & = a_{0} \cdot 1 + a_{1} \cdot 3.6 + a_{2} \cdot 12.96 = 1.236 \end{aligned}

Do you see a 10 Matrix-Vector Multiplication problem? Do you see columns?

⟹ {[\begin{matrix} 1 & 3.0 & 9.00 \\ 1 & 3.3 & 10.89 \\ 1 & 3.6 & 12.96 \end{matrix}]}_{3 \times 3} {[\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \end{matrix}]}_{3 \times 1} = {[\begin{matrix} 3 \\ 2.559 \\ 1.236 \end{matrix}]}_{3 \times 1}

First two columns are linearly independent.
It is also not possible to create column 3 by multiplying a scalar against column 2.
& Note that in a quadratic equation, the columns are always linearly independent.

Remark.
In it's current form, our linear-systems problem is difficult to solve. Instead of attempting to solve it how it currently is, let's transform $A \vec{x} = \vec{b}$ into the system $U \vec{x} = \vec{y}$ where $U$ is an 8.9 Upper-Triangular Matrix so we can use 12.2 Backwards Substitution.
$⧫$

\begin{proof}[Solution.]

Step 1: Identify first pivot

We want to identify the first entry with row and column index 1. This entry must be nonzero and will be referred to as the first pivot of our matrix. Likewise, column 1 will be referred to as the first pivot column.

For our matrix, $a_{11} = 1$ is nonzero. Thus, we will make this our first pivot.

[\begin{matrix} 1 & 3.0 & 9.00 \\ 1 & 3.3 & 10.89 \\ 1 & 3.6 & 12.96 \end{matrix}]

Column $A (:, 1)$ of matrix $A$ is the first pivot column.

Step 2: Create zeroes in all entries below the first pivot

To turn our matrix into an 8.9 Upper-Triangular Matrix, we can multiply the original system of equation by a sequence of shear matrices to introduce zeroes in all entries to our pivot column $A (:, 1)$ that are under the pivot.

In our case, we need to annihilate entries $a_{21}$ and $a_{31}$ .

[\begin{matrix} 1 & 3.0 & 9.00 \\ 1 & 3.3 & 10.89 \\ 1 & 3.6 & 12.96 \end{matrix}]

First, let's transform $a_{21}$ to zero. How can we do this via 6.1 Linear Combination of Vectors?

\begin{aligned} - 1 A (1, :) + A (2, :) & = - 1 [\begin{array}{c} 1 & 3.0 & 9.00 \end{array}] + 1 [\begin{array}{c} 1 & 3.3 & 10.89 \end{array}] \\ = [\begin{array}{c} 0 & 0.3 & 1.89 \end{array}] \end{aligned}

When modifying row 2, we do not want to touch rows 1 or 3. We can do this by multiplying $A$ on the left-hand side by a shear matrix $S_{i k} (c)$ .

& Recall in 11 Matrix-Matrix Multiplication, we can introduce a zero by adding a scalar multiple of row 1 to row 2.
Why do we use a shear matrix to introduce zeroes? #question
- Recall 10.2 Matrix-Column-Vector Multiplication via Dot Products

We need to find what our $c$ value will be for our shear matrix.

\begin{aligned} S_{31} (c) \cdot A & = [\begin{array}{c} \times & \times & \times \\ \times & \times & \times \\ 0 & \times & \times \end{array}] \\ ⟹ {Entry}_{31} (S_{31} (c) \cdot A) & = 0 \\ ⟹ {Row}_{3} (S_{31} (c) \cdot {Column}_{1} (A) & = 0 \\ ⟹ [\begin{array}{c} c & 0 & 1 \end{array}] [\begin{array}{c} 1 \\ 1 \\ 1 \end{array}] & = c \cdot 1 + 1 \cdot 1 = 0 \\ ⟹ c & = - \frac{1}{1} \end{aligned}

Notice that our $c$ value is negative. We make this value negative to cancel out our entry in $A$ .
For our shear matrix, we let $c = - \frac{a_{21}}{a_{11}} = - 1$ . Thus, we multiply $A$ on the left by $S_{21} (- 1)$ :

\begin{aligned} [\begin{array}{c} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 3.0 & 9.00 \\ 1 & 3.3 & 10.89 \\ 1 & 3.6 & 12.96 \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] & = [\begin{array}{c} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{c} 3.00 \\ 2.559 \\ 1.236 \end{array}] \\ [\begin{array}{c} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 1 & 3.6 & 12.96 \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] & = [\begin{array}{c} 3.00 \\ - 0.442 \\ 1.236 \end{array}] \end{aligned}

Because we are working with an equation, anything done to the left of the $=$ sign has to be done on the right also.

Next, let's create a zero at $a_{31} = 1$ by taking the linear combination $- 1 A (1, l :) + A (3, :)$ .

- 1 [\begin{matrix} 1 & 3.00 & 9.00 \end{matrix}] + 1 [\begin{matrix} 1 & 3.6 & 12.96 \end{matrix}] = [\begin{matrix} 0 & 0.6 & 3.96 \end{matrix}]

Again, we do not touch the other rows. We will multiply on the left by shear matrix $S_{31} (- 1)$ where $k = 1$ and $i = 3$ because we add a scalar multiple of row 1 to row 3 (AKA linear combination). We set $c = - \frac{a_{31}}{a_{11}} = - 1$ and multiply:

\begin{aligned} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 1 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 1 & 3.6 & 12.96 \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 1 & 0 & 1 \end{array}] [\begin{array}{c} 3.00 \\ - 0.442 \\ 1.236 \end{array}] \\ [\begin{array}{c} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.6 & 3.96 \end{array}] & = [\begin{array}{c} 3.00 \\ - 0.442 \\ - 1.764 \end{array}] \end{aligned}

Step 3: Identify next pivot

To continue transforming our matrix into an 8.9 Upper-Triangular Matrix, we move to the next row down and next column to the right. The first nonzero entry in this row is the second pivot, and this column is called the second pivot column.

In our case, our second pivot is at entry $(2, 2)$ .

[\begin{matrix} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.6 & 3.96 \end{matrix}]

Column $A (:, 2)$ is our second pivot column.

Does it matter what pivot we choose?

When we think about constructing an upper-triangular matrix, we generally prefer to choose our pivots along the diagonal because it is easier to keep track of. In much more complex situations like computer memory optimization, it may be better to choose a pivot that is not along the diagonal.

Step 4: Create zeros in all entries below current pivot

Again, we use a series of shear matrices to introduce zeros in all entries under the current pivot in our current pivot column.

Let's identify where we need to introduce zeros.

[\begin{matrix} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.6 & 3.96 \end{matrix}]

Because we want to eliminate the entry at row 3 column 2 of $A$ , we choose $i = 3$ and $k = 2$ for our shear matrix. Let's find our $c$ value that we will multiply by.

\begin{aligned} S_{32} (c) \cdot A & = [\begin{array}{c} \times & \times & \times \\ \times & \times & \times \\ \times & 0 & \times \end{array}] \\ ⟹ {Entry}_{32} (S_{32} (c) \cdot A) & = 0 \\ ⟹ {Row}_{3} (S_{32} (c)) \cdot {Column}_{2} (A) & = 0 \\ ⟹ [\begin{array}{c} 0 & c & 1 \end{array}] [\begin{array}{c} 3.0 \\ 0.3 \\ 0.6 \end{array}] & = c \cdot 0.3 + 1 \cdot 0.6 = 0 \\ ⟹ c & = \frac{0.6}{0.3} \\ ⟹ c & = - 2 \end{aligned}

Now that we have our shear matrix $S_{32} (- 2)$ , let's multiply to find:

\begin{aligned} [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 2 & 1 \end{array}] [\begin{array}{c} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.6 & 3.96 \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & - 2 & 1 \end{array}] [\begin{array}{c} 3.00 \\ - 0.442 \\ - 1.764 \end{array}] \\ [\begin{array}{c} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.0 & 0.18 \end{array}] & = [\begin{array}{c} 3.00 \\ - 0.442 \\ - 0.88 \end{array}] \end{aligned}

Step 5: Repeat until all sub-diagonal elements are zero

If matrix $A$ is not fully transformed into an upper-triangular coefficient matrix $U$ , repeat steps 3 and 4.

In our gravity model, we are already done and do not need to repeat steps 3 and 4.

\underset{U}{\underset{⏟}{[\begin{matrix} 1 & 3.0 & 9.00 \\ 0 & 0.3 & 1.89 \\ 0 & 0.0 & 0.18 \end{matrix}]}} \underset{\vec{x}}{\underset{⏟}{[\begin{matrix} a_{0} \\ a_{1} \\ a_{2} \end{matrix}]}} = \underset{\vec{y}}{\underset{⏟}{[\begin{matrix} 3.00 \\ - 0.442 \\ - 0.880 \end{matrix}]}}

Now, we have transformed our original matrix $A$ into our upper triangular matrix $U$ . If we set $E = S_{32} (- 2) \cdot S_{31} (- 1) \cdot S_{21} (- 1)$ , we can write a more simplified version of our system:

\begin{aligned} E A = U, & E \vec{b} = \vec{y} \end{aligned}

Now, our system is designed so any solution of $U \vec{x} = \vec{y}$ will also be a solution to our original linear system of equations $A \vec{x} = \vec{b}$ . We can now solve for the values of $\vec{x}$ row-by-row using 12.2 Backwards Substitution.

0.18 a_{2} = - 0.882 ⟹ a_{2} = - \frac{0.882}{0.18} = - 4.9

Using our linear equation result from row 3, we see

0.3 a_{1} + 1.89 a_{2} = - 0.442 ⟹ a_{1} = \frac{- 0.442 - 1.89 \cdot (- 4.9)}{0.3} = 29.4

Finally, we find $a_{0}$ using our previous values

1 a_{0} + 3 a_{1} + 9 a_{2} = 3 ⟹ a_{0} = \frac{3 - 3 \cdot 29.4 - 9 \cdot (- 4.9)}{1} = - 41.1

Plugging into our original model $h (t)$ , we see

h (t) = - 4.9 t^{2} + 29.4 t - 41.1

We can simplify our quadratic polynomial by completing the square.

\begin{aligned} h (t) & = - 4.9 t^{2} + 29.4 t - 41.1 \\ = - 4.9 (t^{2} + 6 t + 9 - 9) - 41.1 \\ = - 4.9 (t - 3)^{2} + 3 \end{aligned}

Our model indicates that we started our experiment at $t = 3$ seconds dropping from an initial height of $3$ meters with $0$ initial velocity. Thus, we can conclude our model $h (t)$ is valid for the domain $[3, 3.782)$ .

! Our model EXACTLY matches our collected data. In real life, this is very rare as there are many ways to introduce errors into data.

Remark. This example demonstrates regular Gaussian Elimination to solve a system of $n$ equations with $n$ unknowns. This technique works well for a special type of matrix called a 12.3 Regular Matrix.
$⧫$
\end{proof}