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\homework{3: Orthogonality}{Due Monday, January 27, 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!  If you have any questions, feel free to contact either Yihan or I via email or office hours.  Have fun!

\begin{enumerate}
\item Find an orthonormal basis for the space
\begin{align*}
V = \{ (w,x,y,z) ~|~ w + x + y + z = 0\}.
\end{align*}
\item \begin{enumerate}
\item Find the three distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$ for the matrix $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 3\\ 0 & 3 &2\end{bmatrix}$, and their associated eigenvectors $\vec{v_1}, \vec{v_2}, \vec{v_3}$.
\item Set $\vec{r_1}  = \vec{v_1} / ||\vec{v_1}||, \vec{r_2}  = \vec{v_2} / ||\vec{v_2}||, \vec{r_3}  = \vec{v_3} / ||\vec{v_3}||.$  Prove that the set $R = \{\vec{r_1}, \vec{r_2}, \vec{r_3}\}$ is an orthonormal basis for $\mathbb{R}^3$.
\item Write $A$ as a matrix under the basis $R$.  What kind of matrix is $A_R$?
\end{enumerate}
\item \begin{enumerate}
\item Can you find four nonzero vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ such that 
\begin{multicols}{2}
\begin{itemize}
\item $\vec{a}$ is orthogonal to $\vec{b}$,
\item $\vec{b}$ is orthogonal to $\vec{c}$,
\item $\vec{c}$ is orthogonal to $\vec{d}$,
\item $\vec{d}$ is orthogonal to $\vec{a}$,
\end{itemize}
\end{multicols}
but $\vec{a}$ is \textbf{not} orthogonal to $\vec{c}$?  Either find such a set of four vectors and demonstrate they have these properties, or prove no such set can exist.
\item Can you find five nonzero vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}, \vec{e} \in \mathbb{R}^2$, such that 
\begin{multicols}{2}
\begin{itemize}
\item $\vec{a}$ is orthogonal to $\vec{b}$,
\item $\vec{b}$ is orthogonal to $\vec{c}$,
\item $\vec{e}$ is orthogonal to $\vec{a}$?
\item $\vec{c}$ is orthogonal to $\vec{d}$,
\item $\vec{d}$ is orthogonal to $\vec{e}$, and
\end{itemize}
\end{multicols}
\end{enumerate}
\item In class, our definition of \textbf{orthogonality} came from our idea of what a \textbf{dot product} was!  Surprisingly, however, there are many different notions of dot product.  Consider the following definition: for $A = $ the matrix $\begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix}$, and any two vectors $\vec{a}, \vec{b} \in \mathbb{R}^2$, define the \textbf{inner product of $\vec{x}$ and $\vec{y}$ with respect to $A$}, $\langle \vec{a}, \vec{b} \rangle_A$, as follows:
\begin{align*}
\langle \vec{a}, \vec{b} \rangle_A = \left( A \cdot \vec{a} \right) \cdot \vec{b}.
\end{align*}
For example, the inner product of $(1,2), (3,4)$ is
\begin{align*}
\langle (1,2), (3,4)\rangle_A = \left(\begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix} \cdot \begin{bmatrix} 1\\2\end{bmatrix}\right) \cdot (3,4) = (4,5) \cdot (3,4) = 32.
\end{align*}
We can use this inner product to define a ``new'' notion of orthogonality: we say that two vectors $\vec{a}, \vec{b} \in \mathbb{R}^2$ are \textbf{orthogonal with respect to the inner product given by $A$} if $\langle \vec{a}, \vec{b} \rangle_A$ is 0.
\begin{enumerate}
\item Find six distinct nonzero vectors $\vec{a_1}, \ldots \vec{a_6}$ in $\mathbb{R}^2$, none of which are scalar multiples of each other, such that the inner products $\langle \vec{a_1}, \vec{a_2} \rangle_A, \langle \vec{a_3}, \vec{a_4} \rangle_A, \langle \vec{a_5}, \vec{a_6} \rangle_A$ are all zero.  
\item Under the normal notion of dot product and orthogonality, two vectors were orthogonal if the angle between them was a right angle.  Is that true for the inner product with respect to $A$?  Why?
\item Suppose I told you that two nonzero vectors $\vec{a}, \vec{b}$ met at an angle greater than $120^\circ$.  Is it possible that $\langle \vec{a}, \vec{b} \rangle_A = 0$?  Either prove that no such pair can exist, or create a pair that meet at such an angle and have dot product equal to 0.
\end{enumerate}


\item In Math 108A, we ran into the idea of \textbf{polynomials} as vector spaces.  Specifically, consider the set $\mathcal{P}_3(\mathbb{R})$, the collection of all polynomials of degree 3 or less: i.e.\
\begin{align*}
\mathcal{P}_3(\mathbb{R}) = \{ a + bx + cx^2 + dx^3 ~|~ a,b,c,d \in \mathbb{R}\}.
\end{align*}
This is a vector space!  Consider the following notion of ``dot product'' for these polynomials: for any pair of polynomials $p(x), q(x) \in \mathcal{P}_3(\mathbb{R})$, we define their \textbf{inner product}, $\langle p(x), q(x)\rangle$, as follows:
\begin{align*}
\langle p(x), q(x) \rangle = \int_{-1}^1 p(x)\cdot q(x) ~dx.
\end{align*}
For example, the inner product of $1+x$ and $x + x^2$ is the quantity
\begin{align*}
\langle 1+x, x+x^2 \rangle = \int_{-1}^1 (1+x)(x+x^2)~dx &=  \int_{-1}^1 x + 2x^2 + x^3 ~dx \\
&= \left(\frac{x^2}{2} + \frac{2x^3}{3} + \frac{x^4}{4} \right) \Bigg|_{-1}^1 = \frac{4}{3}
\end{align*}
We can use this inner product to define a notion of \textbf{orthogonality}: we say that two polynomials $p(x), q(x)$ are orthogonal if and only if their inner product $\langle p(x), q(x) \rangle$ is 0. 
\begin{enumerate}
\item Warmup: choose four nonzero polynomials $p_1(x), q_1(x), p_2(x), q_2(x)$, and calculate the dot products $\langle p_1(x), q_1(x) \rangle, \langle p_2(x), q_2(x) \rangle$.
\item Find a basis for $\mathcal{P}_3(\mathbb{R})$ that is made of orthogonal polynomials.  (Note: because $\mathcal{P}_3(\mathbb{R})$ is a four-dimensional space, as you have four degrees of freedom in the polynomial $a + bx + cx^2 + dx^3$, you should get four polynomials in your basis.)
\end{enumerate}
\end{enumerate}





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