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Problem 1: Let $(\mathbb{R}^{4 \times 1},\langle \vec{x},\vec{y} \rangle)$ be an inner product space where $\langle \vec{x},\vec{y} \rangle$ denotes dot product. Show that the vectors $\vec{a}=\begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} -4 \\ -3 \\ 2 \\ 1 \end{bmatrix}$ are orthogonal vectors.
Solution: Compute $$\begin{array}{ll} \left\langle \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}, \begin{bmatrix} -4 \\ -3 \\ 2 \\ 1 \end{bmatrix} \right\rangle &= (1)(-4) + (2)(-3) + (3)(2) + (4)(1) \\ &=-4 - 6 + 6 + 4 \\ &= 0, \end{array}$$ showing the vectors are orthogonal.
Problem 5: Consider the vector space $(\mathbb{P},\langle \cdot,\cdot \rangle)$ where the inner product is given by $$\langle p(x),q(x) \rangle = \displaystyle\int_{-\infty}^{\infty} p(x)q(x)e^{-x^2} dx.$$ It can be shown (via methods of problem 4) that the moments in this inner product space are $$\langle 1,1 \rangle=\sqrt{\pi},$$ $$\langle x,1 \rangle=0,$$ $$\langle x^2,1 \rangle=\dfrac{\sqrt{\pi}}{2},$$ $$\langle x^3,1 \rangle = 0,$$ $$\langle x^4,1 \rangle = \dfrac{3\sqrt{\pi}}{4},$$ $$\langle x^5,1 \rangle = 0,$$ $$\langle x^6,1 \rangle = \dfrac{15\sqrt{\pi}}{8}.$$ Use these moments and the "linear in the first argument" property of inner products (noted here) to compute $\langle 4x^2+3x+9,1\rangle$ and $\langle 32x^5-64x^3+24x,1\rangle$.
Solution: Recall that the "linear in the first argument" says that $$\langle \alpha \vec{v} + \beta \vec{u}, \vec{w} \rangle = \alpha \langle \vec{v}, \vec{w} \rangle + \beta \langle \vec{u}, \vec{w} \rangle.$$ Hence we may compute $$\begin{array}{ll} \langle 4x^2+3x+9,1 \rangle &= 4 \langle x^2,1 \rangle + 3 \langle x, 1 \rangle + 9 \langle 1, 1 \rangle \\ &=4 \dfrac{\sqrt{\pi}}{2} + 0 + 9 \sqrt{\pi} \\ &=11 \sqrt{\pi} \end{array}$$ and $$\begin{array}{ll} \langle 32x^5-64x^3+24x,1 \rangle &= 32 \langle x^5,1 \rangle - 64 \langle x^3,1 \rangle + 24 \langle x,1 \rangle \\ &=0 + 0 + 0 \\ &=0. \end{array}$$