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\homework{2: Changes of Basis}{Due Wednesday, January 22, by 1:30pm}


Homework problems need to show work and contain proofs in order to receive full credit.  Simply stating an answer is only half of the problem in mathematics; you also need to include an argument that persuades your audience that your answer is correct!  As always, if you have any questions, feel free to contact either Yihan or I via email or office hours.  Have fun!

\begin{enumerate}


\item \begin{enumerate}
\item Take the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by 
\begin{align*}
T(x,y,z) = (x+y, y+z, x+z).
\end{align*}
 Write it as a matrix using the basis 
\begin{align*}
B = \{(1,0,0), (1,1,0), (1,1,1)\}.
\end{align*}
\item Take the linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^4$ defined by 
\begin{align*}
T(w,x,y,z) = (w+x+y, z, w-x-y, z).
\end{align*}
  Write it as a matrix using the basis 
\begin{align*}
B = \{(1,1,1,1), (1,1,0,0), (1,-1,0,0), (0,0,1,-1)\}.
\end{align*}
\end{enumerate}

\item \begin{enumerate}
\item Consider the identity transformation $id: \mathbb{R}^n \to \mathbb{R}^n$, that sends every vector to itself: i.e.
\begin{align*}
id(\vec{x}) = \vec{x},
\end{align*}
for any $\vec{x} \in \mathbb{R}^n$.  Prove that for \textbf{any} basis $B = \{\vec{b_1}, \ldots \vec{b_n}\}$, the matrix associated to $id$ in the base $B$ is always the identity matrix: i.e.\
\begin{align*}
id_{\textrm{written as a matrix in base }B} = \begin{bmatrix} 1 & 0 & \ldots & 0\\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0& \ldots & 1\end{bmatrix}_{B.}
\end{align*}

\item In part (b), you showed that the identity transformation is represented as the same matrix under any basis.  Are there any other linear transformations $\mathbb{R}^n \to \mathbb{R}^n$ that have this property --- i.e.\ that are written as the same matrix no matter what base you pick?  Either find one such matrix and show it has this property, or prove no such matrices exist.
\end{enumerate}

\item Let $T$ be the linear transformation $\mathbb{R}^3 \to \mathbb{R}^3$ corresponding to the matrix
\begin{align*}
T_{\textrm{standard}} =  \begin{bmatrix} -7 & 12 & 0\\ -6 & 10 & 0 \\ 0 & 0& 3\end{bmatrix}
\end{align*}
Also, let $B$ denote the basis $\{\vec{b_1} =  (3,2,0), \vec{b_2} = (4,3,0), \vec{b_3} = (0,0,1) \}$ for $\mathbb{R}^3$.
\begin{enumerate}
\item Write $T$ as a matrix $T_B$ under the basis $B$.
\item In class, we took any $T$ with associated matrices $T_{\textrm{standard}}, T_B$ under the standard basis and $B$-basis, and proved that
\begin{align*}
T_{\textrm{standard}} =  \begin{bmatrix} \vdots  & &\vdots \\ \vec{b_1} & \ldots & \vec{b_n} \\ \vdots & & \vdots \end{bmatrix} \cdot T_B \cdot \begin{bmatrix} \vdots  & &\vdots \\ \vec{b_1} & \ldots & \vec{b_n} \\ \vdots & & \vdots \end{bmatrix}^{-1}_.
\end{align*}
Verify this claim for our matrices here: i.e. show that 
\begin{align*}
 \begin{bmatrix} -7 & 12 & 0\\ -6 & 10 & 0 \\ 0 & 0& 3\end{bmatrix} = \begin{bmatrix} \vdots  &\vdots &\vdots \\ \vec{b_1} & \vec{b_2} & \vec{b_3} \\ \vdots &\vdots  & \vdots \end{bmatrix} \cdot T_B \cdot \begin{bmatrix} \vdots  & \vdots &\vdots \\ \vec{b_1} & \vec{b_2} & \vec{b_3} \\ \vdots & \vdots & \vdots \end{bmatrix}^{-1}_.
\end{align*}
\end{enumerate}

\item Consider the linear transformation $T(x,y) = (x+y, y)$.
\begin{enumerate}
\item Write this linear transformation in the standard basis.
\item Show that there are no values of $a,b,c,d,e,f$ such that both of the following conditions hold:  (1) the equation
\begin{align*}
\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix} a & b \\ c & d\end{bmatrix} =\begin{bmatrix} a & b \\ c & d\end{bmatrix} \cdot \begin{bmatrix} e & 0 \\ 0 & f\end{bmatrix}
\end{align*}
holds, and (2) the matrix $\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is also invertible.
\item Using the above two parts, explain why there is no basis in which $T$ can be written as a diagonal matrix\footnote{This might be a little surprising, as we've written several matrices in this form $BDB^{-1}$, for $D$ a diagonal matrix and $B$ a matrix made out of the elements of some basis.  This is because we've been studying examples that are ``well-behaved'' in some sense; this is an early exposure to the kind of ``defective'' matrices that are also out there in the world!}.
\end{enumerate}
\end{enumerate}





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