Find all unit vectors in \(\mathbb R^2\) with respect to this new inner product.
See Homework 7.2.a; this is similar but with a different magnitude. Unit vectors are the ellipse determined by \(1 = 3x^2 + 2y^2\), that is, the set of vectors;
\[\left\{ \begin{pmatrix} \frac{1}{\sqrt 3} \cos \theta \\ \frac{1}{\sqrt 2} \sin \theta \end{pmatrix} \middle| \theta \in [0, 2\pi) \right\}.\]
Find two different orthonormal bases for \(\mathbb R^2\) with respect to this new inner product.
Our solution for (a) says that two vectors determined by angles \(\alpha\) and \(\beta\) have dot product,
\[\frac{3}{\sqrt 9}\cos \alpha \cos \beta + \frac{2}{\sqrt 4}\sin \alpha \sin \beta = \cos\alpha\cos\beta + \sin\alpha \sin \beta\]This means that two of our unit vectors are orthogonal if \(\alpha\) and \(\beta\) form a right angle (why?). So to come up with orthonormal bases we just have to pick an angle, and then another angle which is orthogonal to it:
\[\left\{ \begin{pmatrix} \frac{1}{\sqrt 3} \cos \theta \\ \frac{1}{\sqrt 2} \sin \theta \end{pmatrix}, \begin{pmatrix} \frac{1}{\sqrt 3} \cos (\theta+\pi/2) \\ \frac{1}{\sqrt 2} \sin (\theta+\pi/2) \end{pmatrix} \right\}.\]For example, we get two different bases if we pick \(\theta = 0\) and \(\theta = \pi/4\).
If \(\vec v = \begin{pmatrix} 3 \\ 2 \\ 1\end{pmatrix}\) and \(\vec u = \begin{pmatrix} -1 \\ 0 \\ 1\end{pmatrix}\), find a decomposition
\[\vec v = \vec v^{||} + \vec v^{\perp}\]where \(\vec v^{||}\) is parallel to \(\vec u\) and \(\vec v^{\perp}\) is orthogonal to \(\vec u\).
Gram-Schmidt says that
\[\vec v^{||} = \frac{\vec v \cdot \vec u}{\vec u \cdot \vec u}\vec u = \frac{-3 + 1}{1 + 1}\begin{pmatrix} -1 \\ 0 \\ 1\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix}.\]Then,
\[\vec v^{\perp} = \vec v - \vec v^{||} = \begin{pmatrix} 3 \\ 2 \\ 1\end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -1\end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 2\end{pmatrix}.\]
For two \(m \times n\) matrices \(M, N\) we can define the inner product,
\[\langle M, N \rangle = \mathop{tr}(M^T N).\]Are the vectors,
\[\begin{pmatrix}2&1 \\ -1&3\end{pmatrix} \quad\textrm{and}\quad \begin{pmatrix}-3&0 \\ 0&2\end{pmatrix}\]orthogonal? Explain why or why not.
Yes, because \(\langle M, N \rangle = \mathop{tr}(M^T N) = 0\).