# Selected Exercises #

## 6.B #

1. (a) Suppose $$\theta \in \mathbf{R}$$. Show that $(\cos \theta, \sin \theta), (-\sin \theta, \cos \theta)$ and $(\cos \theta, \sin \theta), (\sin \theta, -\cos \theta)$ are orthonormal bases of $\mathbf{R}^2$.
(b) Show that each orthonormal basis of $\mathbf{R}^2$ is of the form given by one of the two possibilities of part (a).

(a) First, we show that both lists of vectors are orthonormal (using the Euclidean inner product), i.e. that the inner product of the two vectors equals 0 and that the inner product of each with itself equals 1.

The former,

\begin{aligned} \langle (\cos \theta, \sin \theta), (-\sin \theta, \cos \theta) \rangle & = & \\ -\cos \theta \sin \theta + \sin \theta \cos \theta & = & 0. \\ \end{aligned}

The latter for the first list, \begin{aligned} \langle (\cos \theta, \sin \theta), (\cos \theta, \sin \theta) \rangle & = \\ \cos^2 \theta + \sin^2 \theta & = 1, \\ \end{aligned} and the second,

\begin{aligned} \langle (-\sin \theta, \cos \theta), (-\sin \theta, \cos \theta) \rangle & = & \\ \sin^2 \theta + \cos^2 \theta & = & 1. \\ \end{aligned}

Thus, the two lists are orthonormal. By 6.28, these lists are bases of $\mathbf{R}^2$, since they both have length $2 = \dim\ \mathbf{R}^2$.

(b) We know that, if $u, v$ are nonzero vectors in $R 2. Suppose$ e_1, \dots. e_m $is an orthonormal list of vectors in$V$. Let$ v \in V $, Prove that $$\lVert v \rVert^2 = \lvert \langle v, e_1 \rangle \rvert^2 + \cdots + \lvert \langle v, e_m \rangle \rvert^2$$ if and only if$v \in \operatorname{span}(e_1, \dots, e_m)$. Let’s start with the backwards direction because it’s easier and I’m a lazy programmer at heart. Assume$ v \in \operatorname{span}(e_1, \dots, e_m) $. Then, there exists some list of scalars$a_1, \dots, a_m$such that $$v = a_1 e_1 + \cdots a_m e_m.$$ Similar to the proof of 6.30, we can take the inner product of both sides for each$ e_j $for$j \in 1, \dots, m $and get $$\langle v, e_j \rangle = \langle a_j \rangle,$$ so $$v = \langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m.$$ Then, taking the norm of both sides and using 6.25 gives us $$\lVert v \rVert^2 = \lvert \langle v, e_1 \rangle \rvert^2 + \cdots + \lvert \langle v, e_m \rangle \rvert^2.$$ The other direction of the if-and-only-if implication is trickier… Let $$\xi = v - (\langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m).$$ We aim to show that$ \xi = 0 $. If we take the inner product of both sides with$e_j$for all$j \in 1, \dots, m$, we get $$\langle \xi, e_j \rangle = \langle v, e_j \rangle - \langle v, e_j \rangle = 0.$$ This implies that $$\langle \xi, e_1 \rangle e_1 + \cdots \langle \xi, e_m \rangle e_m = 0,$$ which, by 6.25, implies that $$\lVert \langle \xi, e_1 \rangle e_1 + \cdots \langle \xi, e_m \rangle e_m \rVert^2 = 0.$$ So by 6.13 and 6.25, we have $$\begin{eqnarray} \lVert v \rVert^2 &=& \lVert \xi + (\langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m) \rVert^2 \\\ &=& \lVert \xi \rVert^2 + \lVert (\langle v, e_1 \rangle e_1 + \cdots + \langle v, e_m \rangle e_m) \rVert^2 \\\ &=& \lVert \xi \rVert^2 + \lvert \langle v, e_1 \rangle e_1 \rvert^2 + \cdots + \lvert \langle v, e_m \rangle e_m \rvert^2. \end{eqnarray}$$ Thus,$ \lVert \xi \rVert^2 = 0 $, so$ \xi = 0 $.$ \blacksquare $3. Suppose$ T \in \mathcal{L}(\mathbf{R}^3) $has an upper-triangular matrix with respect to the basis$ (1, 0, 0), (1, 1, 1), (1, 1, 2) $. Find an orthonormal basis of$ \mathbf{R}^3 $(use the usual inner product on$ \mathbf{R}^3 $) with respect to which$ T $has an upper-triangular matrix. We use the Gram-Schmidt procedure to convert$ (1, 0, 0), (1, 1, 1), (1, 1, 2) $to an orthonormal basis in$ V $. For$ e_1 $, we get $$e_1 = \frac{(1, 0, 0)}{\sqrt{1}} = (1, 0, 0)$$. For$e_2$, our numerator will be $$(1, 1, 1) - \langle (1, 0, 0), (1, 0, 0) \rangle (1, 0, 0) = (0, 1, 1)$$ and our denominator will be $$\lVert (0, 1, 1) \rVert = \sqrt{2},$$ giving us $$e_2 = \frac{1}{\sqrt{2}} (0, 1, 1).$$ For$ e_3 $, our numerator will be $$\begin{eqnarray} (1, 1, 2) - \langle (1, 1, 2), \frac{1}{\sqrt{2}} (0, 1, 1) \rangle \frac{1}{\sqrt{2}} (0, 1, 1) - \langle (1, 1, 2), (1, 0, 0) \rangle (1, 0, 0) &=& (1, 1, 2) - (1, \frac{3}{2}, \frac{3}{2}) \\\\ &=& (0, -\frac{1}{2}, \frac{1}{2}), \end{eqnarray}$$ and our denominator will be $$\sqrt{\frac{2}{4}} = \frac{\sqrt{2}}{2},$$ giving us $$e_3 = \frac{1}{\sqrt{2}}(0, -1, 1).$$ Thus, our orthonormal basis is$ (1, 0, 0), \frac{1}{\sqrt{2}} (0, 1, 1), \frac{1}{\sqrt{2}}(0, -1, 1) \$.