Lab 1 Report.tex

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\title{Lab 1 Report}
\author{Isaac Shaeffer and Satori Kubo}

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\begin{document}
\maketitle

\begin{enumerate}
% 1a - 1d
\item 
\begin{enumerate}
% 1a
\item The show() function renders uses formated text or Latex to render output.
% 1b
\item Changing the ring from QQ to RR forces the data in the matrix $A$ to be treated as floating point values. Changing the ring from QQ to RDF forces the data to take on a single decimal point.
% 1c
\item $ B \sim  \left[\begin{array}{rrr} 1 & 0 & \frac{21}{2} \\ 0 & 1 & -\frac{1}{2} \end{array} \right] $
% 1d
\item The command \verb|I4 = identity_matrix(4)| assigns the \( \mathbb{R}^4 \) identity matrix to the variable \verb|I4|. Changing the number 4 to some other integer, say n, assigns the identity matrix for $ \mathbb{R}^n $ to a variable.
\end{enumerate}

% 2a - 2d
\item
\begin{enumerate}
% 2a

\item Assuming we are using the matrices $A$ and $B$ defined in part 1, we can use matrix addition or subtraction if we restrict the number of columns of $B$ to match the number columns in $A$ with the command \verb|matrix_from_columns([0,1])|. Note that in the argument \verb|[0,1]|, the number 0 is referring to the first column of B and 1 is referring to the second column. The following code
\begin{lstlisting}[language=Python]
A = matrix(QQ,[[1, 2],[3,4]])
B = matrix(QQ, [[0, 2, -1],[1, 1, 10]])

C = A + B.matrix_from_columns([0,1])
show(C)
\end{lstlisting}
generates the output $\left[\begin{array}{rr} 1 & 4 \\4 & 5 \end{array}\right].$

\item The command \verb|B.transpose()| creates a new matrix $B^T$ that uses of the rows of B as its columns. That is
$$
B \sim \left[\begin{array}{rrr}
0 & 2 & -1 \\
1 & 1 & 10
\end{array}\right]
\text{ and } B^T \sim
\left[\begin{array}{rr}
0 & 1 \\
2 & 1 \\
-1 & 10
\end{array}\right].
$$

\item
\begin{enumerate}

% 2. (c) i.
\item Does $(AB)^T = A^TB^T$ for any two matrices $A, B$ where $AB$ is defined?
No, since there exist two matrices $A, B$ where $AB$ is defined  but $(AB)^T \neq A^TB^T$. \par 
For example, let $A = \left(\begin{array}{rr}1 & 1 \\1 & 1\end{array}\right) \text{ and } B = \left[\begin{array}{rr}-2 & -3 \\0 & 7\end{array}\right]$. Then
$$
(AB)^T = \left[\begin{array}{rr}
-2 & -2 \\
4 & 4
\end{array}\right]
\text{ and } A^TB^T = 
\left[\begin{array}{rr}
-5 & 7 \\
-5 & 7
\end{array}\right].
$$

Thus $(AB)^T \neq A^TB^T$.

% 2. (c) ii.

\item Does $(A+B)^T =A^T +B^T$ for any two matrices A,B where $A+B$ is defined? \par

Yes, since by Theorem 2.3 part b., we have $(A+B)^T =A^T +B^T$ for any two matrices $A,B$ where $A+B$ is defined.

% 2. (c) iii. 

\item Does $(cA)^T = cA^T$ for any matrix A and scalar c? \par

Yes, since by Theorem 2.3 part c., we have $(rA)^T = rA^T$ for a matrix A and any scalar r.
\end{enumerate}

% 2. (d)
\item

\begin{enumerate}
% 2. (d) i.
\item What happens when you use the inverse command on matrix A from lab task 1? Does it match the formula for a $2 \times 2$ matrix inverses given in section 2.1? \par

Using \verb|A.inverse()| gives us the the matrix  $\left[\begin{array}{rr} -2 & 1 \\\ \frac{3}{2} & -\frac{1}{2} \end{array}\right]$. The inverse formula from section 2.1 gives us the same matrix. So the formula matches what we get from the inverse command.

% 2. (d) ii.
\item What happens when you use the determinant command on matrix A from lab task 1? Does it match the formula for a $2 \times 2$ matrix determinant given in exam 1? \par

Using \verb|A.det()| returns $-2$. Since $det(A) = 4\cdot 1 - 2\cdot 3 = -2$,  \verb|A.det()| matches the formula for a $2 \times 2$ matrix given in exam 1.

% 2. (d) iii.
\item What happens when you use the inverse command on matrix B from lab task 1? Explain what’s happening here? \par

Using \verb|B.inverse()| returns the error message $$\texttt{ArithmeticError: self must be a square matrix}.$$ This error occurs because inverse matrices are only defined for square matrices and $B$ is not a square matrix.


% 2. (d) iv.
\item Define a $2 \times 2$ matrix whose determinant is zero. What happens when you try to determine the inverse of this matrix in Sage? \par

We have $det(A) = 0$ when $D = \displaystyle \left[\begin{array}{rr} 6 & 3 \\ 4 & 2 \end{array} \right]$. Using \verb|D.inverse()| returns the error $$\texttt{ZeroDivisionError: Matrix is singular}$$

% 2. (d) v.
\item Does $(A^T)^{-1} = (A^{-1})^T$, where A is the matrix from lab task 1? In your report, write down the lines of code you used to determine if this is true or not. \par

The relationship $(A^T)^{-1} = (A^{-1})^T$ is true for the matrix $A$ since 
\[
(A^T)^{-1} = 
\left[\begin{array}{rr}
-2 & \frac{3}{2} \\
1 & -\frac{1}{2}
\end{array}\right]
\text{ and }
(A^{-1})^T = 
\left[\begin{array}{rr}
-2 & \frac{3}{2} \\
1 & -\frac{1}{2}
\end{array}\right]
\]

We used the following code in Sage: 
\begin{lstlisting}[language=Python]

A = matrix(QQ,[[1, 2],[3,4]])

(A.transpose()).inverse()
(A.inverse()).transpose()

show("$(A^T)^{-1}$ =", (A.transpose()).inverse()) 
show("$(A^{-1})^T$ =", (A.inverse()).transpose())

\end{lstlisting}


\end{enumerate}

\end{enumerate}

% 3a - 3i.
\item Let's explore the geometry of linear transformation $T: \mathbb{R}^2  \to \mathbb{R}^2 $. Use the code below to define $ M = \displaystyle \left[\begin{array}{rr} -1 & 0 \\ 0 & 1 \end{array} \right] $ and the column vectors, $ \vec{e_1} $ and $ \vec{e_2} $ of the $ 2 \times 2 $ identity matrix.

\begin{enumerate}
% 3a
\item Describe the output of this code. What are the blue arrows compared to the purple arrows? \par

The blue arrows are the vectors, $\vec{e_1}=\verb|[1,0]|$, $\vec{e_2}=\verb|[0,1]|$, $\vec{u_1}=\verb|[1,2]|$, and $\vec{u_2}=\verb|[-2,-1]|$. The purple arrows are the transformed vectors, $M*\vec{e_1}=\verb|[-1,0]|$, $M*\vec{e_2}=\verb|[0,1]|$, $M*\vec{u_1}=\verb|[-1,2]|$, and $M*\vec{u_2}=\verb|[2,-1]|$.

% 3b
\item Consider $M : \mathbb{R}^2 \to \mathbb{R}^2$ as a linear transformation. Describe how $M$ transforms an input vector $\vec{v}$. \par
The geometry of the transformation is a reflection through the ${x_2}$-axis.

% 3c
\item Change matrix $M$ in the code to $M1=\displaystyle \left[\begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]$ and describe how $M1$ transforms vectors in $\mathbb{R}^2$. \par
The geometry of the transformation is a reflection through the line ${x_2}$ = ${x_1}$. \par

% 3d
\item Change matrix $M$ in the code to $M2=\displaystyle \left[\begin{array}{rr} 0 & -1 \\ -1 & 0 \end{array} \right]$ and describe how $M2$ transforms vectors in $\mathbb{R}^2$. \par
The geometry of the transformation is a reflection through the line ${x_2}$ = $-{x_1}$. \par

% 3e
\item Change matrix $M$ in the code to $M3=\displaystyle \left[\begin{array}{rr} 1 & 0 \\ 0 & 0 \end{array} \right]$ and describe how $M3$ transforms vectors in $\mathbb{R}^2$. \par
The geometry of the transformation is a projection onto the ${x_1}$-axis. \par

% 3f
\item Change matrix $M$ in the code to $M4=\displaystyle \left[\begin{array}{rr} 0 & 2 \\ 4 & 0 \end{array} \right]$ and describe how $M4$ transforms vectors in $\mathbb{R}^2$. \par
The geometry of the transformation is a reflection through the line ${x_2}$ = ${x_1}$ and stretching horizontally and vertically by factors of 2 and 4, respectively. \par

% 3g
\item Change the first two lines in the code to the following. Then change the value of $\theta$ (theta) to see what happens to the vectors. Describe what the transformation $R=\displaystyle \left[\begin{array}{rr} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{array} \right]$ does to vectors in $\mathbb{R}^2$. \par
The geometry of the transformation is a clockwise rotation around the origin by $\theta$ radians.

% 3h
%\item Compute $M2*R$ and describe the transformation it does to vectors in $\mathbb{R}^2$. Then do the same for $R*M2$. \par
%$M2*R = \left(\begin{array}{rr}\sin\left(\theta\right) & -\cos\left(\theta\right) \\-\cos\left(\theta\right) & -\sin\left(\theta\right)\end{array}\right)$; 
%therefore, the geometry of the transformation is a reflection through the ${x_1}$-axis. Also, $R*M2 = \displaystyle \left[\begin{array}{rr} -1 & 0 \\ 0 & 1 \end{array} \right]$; therefore, the geometry of the transformation is a reflection through the ${x_2}$-axis. \par

% 3. (h) redo
\item Compute $M2*R$ and describe the transformation it does to vectors in $\mathbb{R}^2$. Then do the same for $R*M2$. \par

The matrix $M2*R = \left[\begin{array}{rr} \sin\left(\theta\right) & -\cos\left(\theta\right) \\ -\cos\left(\theta\right) & -\sin\left(\theta\right) \end{array}\right]$. \par 
This transformation rotates a vector clockwise $\theta$ radians about the origin and then reflects the rotated vector through the line $x_2 = -x_1$.  

The matrix $R*M2 = \left[\begin{array}{rr} -\sin\left(\theta\right) & -\cos\left(\theta\right) \\ -\cos\left(\theta\right) & \sin\left(\theta\right) \end{array}\right]$. \par
This transformation first reflects a vector through the line $x_2 = -x_1$ and then rotates the reflected vector clockwise $\theta$ radians about the origin.


% 3i
\item For $\theta=\frac{\pi}{2}$, compute $R^2=R*R$ and describe the transformation it does to vectors in $\mathbb{R}^2$. \par
$R^2=R*R= \left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] \left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right] = \left [\begin {array}{rr} -1 & 0 \\ 0 & -1 \end {array} \right]$. Therefore, the geometry of the transformation is a reflection through the origin. \par


\end{enumerate}

% 4
\item
\begin{enumerate}
\item Let $I_2 = \left[ \vec{e}_1 \text{ } \vec{e}_2\right] = \left[\begin{array}{rr}1 & 0 \\ 0 & 1\end{array}\right]$ , let  $T: \mathbb{R}^2 \to \mathbb{R}^2 \text{ be the transformation } T\left(  \left[ \begin{array}{r} a \\ b \end{array} \right] \right) = \left[ \begin{array}{r} a - b \\ a+ b \end{array} \right]$ and let $A$ be the standard matrix for T. \par

First, we will show that $T$ is linear. Thus,
\[
\begin{split}
T\left( \left[ \begin{array}{r} a  \\ 0  \end{array} \right]  + \left[ \begin{array}{r} 0  \\ b  \end{array} \right]\right) &= T\left(  \left[ \begin{array}{r} a \\ b \end{array} \right] \right) \\ 
&= \left[ \begin{array}{r} a - b \\ a + b \end{array} \right] \\
&= \left[ \begin{array}{r} a - 0  \\ a + 0  \end{array} \right] + \left[ \begin{array}{r} 0 - b  \\ 0 + b  \end{array} \right] \\
&= T\left(\left[ \begin{array}{r} a  \\ 0  \end{array} \right]   \right) +  T\left(\left[ \begin{array}{r} 0  \\ b  \end{array} \right]  \right). \\
\end{split}
\]

Now we will find the standard matrix of T. Thus

\[
\begin{split}
T(\vec{e}_1 ) 
&= T\left(\left[ \begin{array}{r} 1  \\ 0  \end{array} \right]   \right)
= \left[ \begin{array}{r} 1 - 0 \\ 1 + 0 \end{array} \right]
= \left[ \begin{array}{r} 1 \\ 1 \end{array} \right] \text{ and }\\
T(\vec{e}_2) 
&= T\left(\left[ \begin{array}{r} 0  \\ 1  \end{array} \right]   \right)
= \left[ \begin{array}{r} 0 - 1 \\ 0 + 1 \end{array} \right]
= \left[ \begin{array}{r} -1 \\ 1 \end{array} \right].
\end{split}
\]
So it follows that \[
A 
= \left[ T(\vec{e_1}) \text{  } T(\vec{e_2}) \right]
= \left[ \begin{array}{rr} 1 & -1 \\ 1 & 1 \end{array} \right].
\]

\item Evaluating the left side changes the standard matrix to $\left(\begin{array}{rr}1 & 1 \\ -1 & 1\end{array}\right)$.

\end{enumerate}

% 5
\item Let  $J: \mathbb{R}^2 \to \mathbb{R}^4$ map 
$\left[ \begin{array}{r} x_1 \\ x_2 \end{array} \right] = \left[ \begin{array}{cc} 2x_1 -5x_1 \\ 0 \\ x_1 + 4x_2 \\ x_2 \end{array} \right].$ \par
Then the columns of the standard matrix for J are
\[
\begin{split}
&J \left( \left[ \begin{array}{r} 1 \\ 0 \end{array} \right] \right) = \left[ \begin{array}{cc} 2\cdot1 -5\cdot0 \\ 0 \\ 1 + 4\cdot0 \\ 0 \end{array} \right] = \left[ \begin{array}{cc} 2 \\ 0 \\ 1 \\ 0 \end{array} \right] \text{ and } \\
&J \left( \left[ \begin{array}{r} 0 \\1 \end{array} \right] \right) = \left[ \begin{array}{cc} 2\cdot0 -5\cdot1 \\ 0 \\ 0 + 4\cdot1 \\ 1 \end{array} \right] = \left[ \begin{array}{rr} -5 \\ 0 \\ 4 \\ 2 \end{array} \right].
\end{split}
\]
Therefore, the standard matrix for J is $ \left[ \begin{array}{rr} 2 & -5 \\ 0 & 0 \\ 1 & 4 \\ 0 & 1 \end{array}\right]$.
\end{enumerate}

\end{document}