Exercises: Inner Product Spaces

Solution. The lengths of given vectors are \[\begin{aligned} \lVert (1,\,2,\,-1)\rVert &= \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} ,\\[4pt] \lVert (2,\,1,\,4)\rVert &= \sqrt{2^2 +1^2 +4^2} = \sqrt{21} . \end{aligned}\] Let \(\theta\) be the angle between two vectors. Since \[\langle (1,\,2,\,-1) \,\vert\, (2,\,1,\,4)\rangle = 2+2-4 = 0,\] we have \[\cos\theta = \frac{0}{\sqrt{6} \sqrt{21}} = 0,\] that is, two vectors are orthogonal.

Solution. Let \(\theta\) be the angle between two vectors. \[\begin{aligned} \cos\theta &= \frac{\langle (1,\,2,\,4) \,\vert\, (2,\,5,\,1)\rangle}{\lVert (1,\,2,\,4)\rVert \lVert (2,\,5,\,-1)\rVert} \\[4pt] &= \frac{16}{\sqrt{21}\sqrt{30}} = \frac{16}{3\sqrt{70}}. \end{aligned}\] Hence \[\theta = \cos^{-1} \frac{16}{3\sqrt{70}} \approx 0.8796 \,(\approx 50.4 ^\circ ) \]

Solution. Since \[(1,\,2,\,0)\times (1,\,0,\,1) = (2,\,-1,\,-2),\] the desired vector is \[k(2,\,-1,\,-2)\] where \(k\) is any real number.

Solution. \[\langle f\,\vert\,g\rangle = \int_{-\pi}^{\pi} 2x \sin x\, dx = 4\pi .\]

Solution. \[\begin{aligned} \langle f\,\vert\,f\rangle &= \int_{-\pi}^{\pi} 2x\cdot 2x \,dx = \frac{8}{3} \pi^3 , \\[4pt] \lVert f\rVert &= \sqrt{\frac{8}{3}\pi^3} ,\\[4pt] \langle g\,\vert\,g\rangle &= \int_{-\pi}^{\pi} \sin x \cdot \sin x \,dx = \pi ,\\[4pt] \lVert g \rVert &= \sqrt{\pi} . \end{aligned}\] Let \(\theta\) be the angle between \(f\) and \(g.\) \[\cos\theta = \frac{\langle f\,\vert\,g\rangle}{\lVert f \rVert \lVert g \rVert} = \frac{4\pi}{\sqrt{\frac{8}{3}\pi^3}\sqrt{\pi}} = \frac{\sqrt{6}}{\pi}.\] Hence we have \[\langle f\,\vert\,g\rangle = 4\pi \le \sqrt{\frac{8}{3}}\pi^2 = \lVert f \rVert \lVert g\rVert,\] and we see that Cauchy-Schwarz inequality holds.

Solution. Since \[\langle 1\,\vert\,x^n \rangle = \int_{-1}^{1} 1\cdot x^n \,dx = \frac{1+(-1)^n}{n+1},\] \(1\) and \(x^n\) are orthogonal to each other if and only if \(n\) is an odd number.

Solution. \[\lVert x^n \rVert = \sqrt{\langle x^n \,\vert\, x^n \rangle}= \sqrt{\frac{2}{2n+1}}.\]

Solution. Take \(\sqrt{x}\) and \(\sqrt{\sin x}\) in \(C^0 ( [ 0 ,\, \frac{\pi}{2} ] ) .\) Then, by Cauchy-Schwarz inequality, we have \[\begin{aligned} \int_{0}^{\pi /2} \sqrt{x}\sqrt{\sin x} \,dx &= \left\langle \sqrt{x} \,\vert\, \sqrt{\sin x}\right\rangle \\[4pt] &\le \left\lVert \sqrt{x} \right\rVert \left\lVert \sqrt{\sin x} \right\rVert \\[4pt] &= \sqrt{\int_{0}^{\pi /2} x \,dx} \cdot \sqrt{\int_0^{\pi /2} \sin x\,dx} \\[4pt] &= \sqrt{\frac{\pi^2}{8}} \cdot \sqrt{1} = \frac{\pi}{2\sqrt{2}}. \end{aligned}\] In fact, \[\begin{aligned} \int_{0}^{\pi/2} \sqrt{x\sin x}\,dx &\approx 1.10839 ,\\[4pt] \frac{\pi}{2\sqrt{2}} & \approx 1.11072 . \end{aligned}\]

Solution. An orthogonal family is linearly independent. An \(n\)-dimensional space cannot contain a linearly independent family of more than \(n\) vectors.

Solution. First, normalize \(v_1 :\) \[\begin{aligned} \lVert v_1 \rVert &= \sqrt{2^2 +1^2} = \sqrt{5}, \\[4pt] u_1 &= \frac{1}{\sqrt{5}} (2,\,1) = \left(\frac{2}{\sqrt{5}} ,\, \frac{1}{\sqrt{5}}\right). \end{aligned}\] Second, find \(u_2 :\) \[\begin{aligned} v_2 - \operatorname{proj}_{v_1} v_2 &= (-1,\,3) - \frac{(2,\,1) \cdot (-1,\,3)}{(2,\,1)\cdot (2,\,1)} (2,\,1) \\[4pt] &= (-1,\,3) - \frac{1}{5} (2,\,1) \\[4pt] &= \left( - \frac{7}{5} ,\, \frac{14}{5}\right), \\[4pt] \left\lvert\left(-\frac{7}{5} ,\, \frac{14}{5}\right)\right\rvert &= \frac{7}{\sqrt{5}} ,\\[4pt] u_2 &= \frac{\sqrt{5}}{7}\left(-\frac{7}{5} ,\, \frac{14}{5}\right) \\[4pt] &= \left(-\frac{1}{\sqrt{5}},\, \frac{2}{\sqrt{5}}\right). \end{aligned}\] Now \(u_1,\) \(u_2\) constitute an orthonormal basis for \(\mathbb{R}^2.\)

Solution. First, normalize \(v_1\): \[\begin{aligned} \lVert v_1 \rVert &= \sqrt{9+16+25} = 5\sqrt{2} ,\\[4pt] u_1 &= \frac{1}{5\sqrt{2}}v_1 = \frac{1}{5\sqrt{2}}(3,\,4,\,5)\\[4pt] &= \left( \frac{3}{5\sqrt{2}} ,\, \frac{4}{5\sqrt{2}} ,\, \frac{1}{\sqrt{2}}\right). \end{aligned}\] Next, find \(u_2\): \[\begin{aligned} v_2 - \operatorname{proj}_{v_1} v_2 &= (1,\,0,\,1) - \frac{8}{50} (3,\,4,\,5) \\[4pt] &= (1,\,0,\,1) - \left( \frac{24}{50} ,\, \frac{32}{50} ,\, \frac{40}{50}\right) \\[4pt] &= \left( \frac{13}{25} ,\, -\frac{16}{25} ,\, \frac{1}{5}\right), \\[4pt] \left\lVert \left( \frac{13}{25} ,\, -\frac{16}{25} ,\, \frac{1}{5}\right) \right\rVert &= \frac{1}{25}\sqrt{169+256+25}\\[4pt] &= \frac{\sqrt{450}}{25} = \frac{3\sqrt{2}}{5},\\[4pt] u_2 &= \frac{5}{3\sqrt{2}} \left(\frac{13}{25} ,\, -\frac{16}{25} ,\, \frac{1}{5}\right) \\[4pt] &= \left( \frac{13}{15\sqrt{2}} ,\, -\frac{16}{15\sqrt{2}} ,\, \frac{1}{3\sqrt{2}} \right). \end{aligned}\] Checkup: \[u_1 \cdot u_2 = \frac{39}{150} - \frac{64}{150} + \frac{1}{6} = 0.\]

Solution. Let \[\begin{aligned} w_1 &= (1,\,0,\,1), \\[4pt] w_2 &= (0,\,1,\,0), \\[4pt] W &= \operatorname{Span}(w_1 ,\,w_2 ),\\[4pt] v &= (6,\,2,\,5). \end{aligned}\] Since \(w_1 \bot w_2,\) the projection of \(v\) onto \(W\) is as follows. \[\begin{aligned} \operatorname{proj}_{w_1} v &= \frac{w_1 \cdot v}{w_1 \cdot w_1} w_1 = \frac{11}{2} w_1 = \left( \frac{11}{2} ,\, 0 ,\, \frac{11}{2}\right) ,\\[4pt] \operatorname{proj}_{w_2} v &= \frac{w_2 \cdot v}{w_2 \cdot w_2} w_2 = \frac{2}{1} w_2 = (0,\,2,\,0), \\[4pt] \operatorname{proj}_W v &= \operatorname{proj}_{W_1} v + \operatorname{proj}_{W_2} v =\left( \frac{11}{2} ,\,2,\, \frac{11}{2}\right). \end{aligned}\] Hence \((\frac{11}{2} ,\,2,\,\frac{11}{2})\) is the closest vector.

Solution. Let \[\begin{aligned} w_1 &= (1,\,1,\,2),\\[4pt] w_2 &= (1,\,1,\,-1),\\[4pt] W &= \operatorname{Span}(w_1,\,w_2),\\[4pt]\ v &= (4,\,5,\,-2). \end{aligned}\] Since \(w_1 \bot w_2,\) the projection of \(v\) onto \(W\) is as follows. \[\begin{aligned} \operatorname{proj}_{w_1}v &= \frac{w_1 \cdot v}{w_1 \cdot w_1} w_1 = \frac{5}{6}(1,\,1,\,2) = \left( \frac{5}{6} ,\, \frac{5}{6} ,\, \frac{10}{6}\right),\\[4pt] \operatorname{proj}_{w_2}v &= \frac{w_2 \cdot v}{w_2 \cdot w_2} w_2 = \frac{11}{3} (1,\,1,\,-1) = \left(\frac{11}{3} ,\, \frac{11}{3} ,\, -\frac{11}{3}\right),\\[4pt] \operatorname{proj}_{W}v &= \operatorname{proj}_{w_1}v+\operatorname{proj}_{w_2}v \\[4pt] &= \left(\frac{27}{6} ,\, \frac{27}{6} ,\, -\frac{6}{3}\right) = \left( \frac{9}{2} ,\,\frac{9}{2} ,\, -2 \right). \end{aligned}\] Hence \((\frac{9}{2} ,\, \frac{9}{2} ,\, -2)\) is the closest vector.

Solution. Let \[\begin{aligned} v_1 &= (2,\,1,\,1),\\[4pt] v_2 &= (0,\,2,\,1),\\[4pt] v &= (2,\,2,\,5),\\[4pt] W &= \operatorname{Span}(v_1 ,\,v_2 ). \end{aligned}\] Since \(v_1\) and \(v_2\) are not orthogonal to each other, we have to find an orthogonal basis for \(W.\) The Gram-Schmidt orthogonalization gives us the following: \[\begin{aligned} \lVert v_1 \rVert &= \sqrt{6} ,\\[4pt] u_1 &= \frac{1}{\sqrt{6}}(2,\,1,\,1), \\[4pt] v_2 - \operatorname{proj}_{v_1}v_2 &= (0,\,2,\,1) - \left( 1,\,\frac{1}{2} ,\, \frac{1}{2}\right) = \left( -1 ,\, \frac{3}{2} ,\, \frac{1}{2}\right) ,\\[4pt] \left\lVert \left( -1 ,\, \frac{3}{2} ,\, \frac{1}{2}\right) \right\rVert &= \sqrt{1+\frac{9}{4} + \frac{1}{4}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2},\\[4pt] u_2 &= \frac{2}{\sqrt{14}}\left(-1,\,\frac{3}{2} ,\,\frac{1}{2}\right) = \left(-\frac{2}{\sqrt{14}} ,\, \frac{3}{\sqrt{14}} ,\, \frac{1}{\sqrt{14}}\right). \end{aligned}\] Hence \(u_1,\) \(u_2\) constitutes an orthonormal basis for \(W.\) Now we find the projection of \(v\) onto \(W\): \[\begin{aligned} \operatorname{proj}_{u_1}v &= (u_1 \cdot v)u_1 \\[4pt] &= \frac{11}{6} (2,\,1,\,1) = \left(\frac{11}{3} ,\, \frac{11}{6} ,\, \frac{11}{6}\right) ,\\[4pt] \operatorname{proj}_{u_2}v &= (u_2 \cdot v)u_2 \\[4pt] &= 1\left(-1,\, \frac{3}{2} ,\, \frac{1}{2}\right) = \left(-1,\, \frac{3}{2} ,\, \frac{1}{2}\right),\\[4pt] \operatorname{proj}_{W}v &= \operatorname{proj}_{u_1}v + \operatorname{proj}_{u_2}v \\[4pt] &= \left(\frac{11}{3} - 1 ,\, \frac{11}{6} + \frac{3}{2} ,\, \frac{11}{6} + \frac{1}{2}\right)\\[4pt] &=\left( \frac{8}{3},\, \frac{10}{3} ,\, \frac{7}{3}\right). \end{aligned}\] Therefore \(\left( \frac{8}{3},\, \frac{10}{3} ,\, \frac{7}{3}\right)\) is the closest vector.

Solution. Keep calm and call Calculus. \[\begin{aligned} \int_{-\pi}^{\pi} 1\,dx &= 2\pi, \\[4pt] \lVert 1 \rVert &= \sqrt{2\pi} ,\\[4pt] \int_{-\pi}^{\pi} (\cos nx)^2 \,dx &= \pi ,\\[4pt] \lVert \cos nx \rVert &= \sqrt{\pi} ,\\[4pt] \int_{-\pi}^{\pi} (\sin nx)^2 \,dx &= \pi ,\\[4pt] \lVert \sin nx \rVert &= \sqrt{\pi} ,\\[4pt] \int_{-\pi}^{\pi} \sin(mx) \cdot \cos(nx) \,dx &=0 \\[4pt] \int_{-\pi}^{\pi} \sin(mx) \cdot \sin(nx) \,dx &=0 \\[4pt] \int_{-\pi}^{\pi} \cos(mx) \cdot \cos(nx) \,dx &=0, \end{aligned}\] where \(n\) and \(m\) are distinct nonnegative integers. Hence the desired orthonormal family is \[\frac{1}{\sqrt{2\pi}},\,\, \frac{1}{\sqrt{\pi}}\cos x ,\,\, \frac{1}{\sqrt{\pi}}\sin x ,\,\, \frac{1}{\sqrt{\pi}}\cos 2x ,\,\, \frac{1}{\sqrt{\pi}}\sin 2x ,\,\, \cdots .\]

Solution. Define \[\begin{aligned} a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx , \\[4pt] a_k &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\cos(kx)\,dx ,\\[4pt] b_k &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(kx)\,dx \end{aligned}\] for \(k=1,\,2,\,\cdots,\,n.\) Then \[\begin{aligned} \operatorname{proj}_W(f) &= \frac{a_0}{2} + a_1 \cos x + b_1 \sin x + a_2 \cos 2x + b_2 \sin 2x + \cdots \\[4pt] &\quad + a_n \cos nx + b_n \sin nx . \end{aligned}\]

Solution. \[\operatorname{proj}_w v = \frac{\langle w\,\vert\,v\rangle}{\langle w\,\vert\,w \rangle} w.\]

Solution. We will find an orthogonal basis \(v_1,\) \(v_2,\) \(v_3\) for \(V=\mathbb{R}^3,\) with \(\operatorname{Span}(v_1)=W.\)

Since \(W=\operatorname{Span}((1,\,1,\,1)),\) let \(v_1 = (1,\,1,\,1)\) and take any \(v_2,\) \(v_3\) so that \(v_1,\) \(v_2,\) \(v_3\) constitutes a linearly independent family. Take \[v_1 = (1,\,1,\,1),\,\, v_2 = (1,\,0,\,0),\,\, v_3 =(0,\,1,\,0).\] It is easily shown that \(v_1,\) \(v_2,\) \(v_3\) is linearly independent. (If this family were not linearly independent, we could choose any other \(v_3 '\) that is not parallel to none of \(v_1,\) \(v_2\) and \(v_3.\)) Hence \(v_1,\) \(v_2,\) \(v_3\) constitutes a basis for \(V.\)

Apply Gram-Schmidt orthogonalization: \[\begin{aligned} u_1 &= \frac{1}{\sqrt{3}}v_1 = \left(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right),\\[4pt] \operatorname{proj}_{u_1}v_2 &= \frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{3},\, \frac{1}{3},\, \frac{1}{3}\right),\\[4pt] v_2 - \operatorname{proj}_{u_1}v_2 &= \left(\frac{2}{3} ,\, -\frac{1}{3} ,\, -\frac{1}{3}\right),\\[4pt] \left\lVert v_2 - \operatorname{proj}_{u_1}v_2\right\rVert &= \frac{\sqrt{6}}{3},\\[4pt] u_2 &= \frac{3}{\sqrt{6}}\left(\frac{2}{3},\, -\frac{1}{3} ,\, -\frac{1}{3}\right) = \left(\frac{2}{\sqrt{6}} ,\, -\frac{1}{\sqrt{6}} ,\, -\frac{1}{\sqrt{6}}\right),\\[4pt] \operatorname{proj}_{u_1} v_3 &= \frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}},\,\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{3} ,\, \frac{1}{3} ,\, \frac{1}{3}\right),\\[4pt] \operatorname{proj}_{u_2}v_3 &= -\frac{1}{\sqrt{6}}\left(\frac{2}{\sqrt{6}} ,\, -\frac{1}{\sqrt{6}} ,\, -\frac{1}{\sqrt{6}}\right) = \left(-\frac{1}{3} ,\, \frac{1}{6} ,\, \frac{1}{6}\right) ,\\[4pt] v_3 - \operatorname{proj}_{u_1}v_3 - \operatorname{proj}_{u_2}v_3 &= (0,\,1,\,0) - \left(\frac{1}{3},\,\frac{1}{3},\,\frac{1}{3}\right) - \left(-\frac{1}{3},\,\frac{1}{6},\,\frac{1}{6}\right) \\[4pt] &= \left(0,\,-\frac{1}{2},\,-\frac{1}{2}\right),\\[4pt] u_3 &= \left(0 ,\, -\frac{1}{\sqrt{2}} ,\, -\frac{1}{\sqrt{2}}\right). \end{aligned}\] Since \(u_1,\) \(u_2,\) \(u_3\) is an orthonormal basis for \(V\) and \(\operatorname{Span}(v_1)=W,\) we have \(W^\bot = \operatorname{Span}(u_2 ,\,u_3),\) that is, \(u_2,\) \(u_3\) consitutes a basis for \(W^\bot.\)

Solution(Alt). Take \[v_1 = (1,\,1,\,1),\,\, v_2 =(a,\,-a,\,0),\,\, v_3 = (0,\, -b ,\,b)\] where \(a\) and \(b\) are nonzero real numbers. (Choosing such vectors \(v_2,\) \(v_3\) is depend on your sense of mathematics.) It is easily shown that \(v_1,\) \(v_2,\) \(v_3\) is a linearly independent family and \[v_1 \cdot v_2 =0 ,\,\, v_1 \cdot v_3 =0,\] that is, \[v_1 \bot v_2 \,\,\text{and}\,\, v_1 \bot v_3 .\] Hence \(v_2,\) \(v_3\) constitutes a basis for \(W^\bot.\)

Solution. If \(W=V,\) then \(\operatorname{proj}_W\) is an identity function; If \(W=\left\{ 0 \right\},\) then \(\operatorname{proj}_W (v)=0\) for any \(v\in V.\) In these cases, \(\operatorname{proj}_W\) is obviously a linearly transformation.

We now consider the cases for \(W\ne V\) and \(W\ne\left\{ 0 \right\}.\)

Let \(u,\,v\in V\) and \(k\in K,\) where \(K\) is the field over which \(V\) is defined. Since \(V = W \oplus W^\bot,\) \(u\) and \(v\) are expressible in the unique forms: \[\begin{aligned} u &= u_1 + u_2 ,\,\, u_1 = \operatorname{proj}_W u \in W ,\,\, u_2 = u-u_1 \in W^\bot ,\\[4pt] v &= v_1 + v_2 ,\,\, v_1 = \operatorname{proj}_W v \in W ,\,\, v_2 = v-v_1 \in W^\bot . \end{aligned}\] We now see that \[u+v = (u_1 + v_1) + (u_2 + v_2);\] Since \(u_1 +v_1 \in W,\) \(u_2 + v_2 \in W^\bot,\) this expression is unique, and we have \[\operatorname{proj}_W(u+v) = u_1 +v_1.\] Hence we have \[\operatorname{proj}_W(u+v) = u_1 +v_1 = \operatorname{proj}_W u + \operatorname{proj}_W v.\] In the same manner, since \[ku = ku_1 + ku_2 ,\,\, ku_1 \in W ,\,\, ku_2 \in W^\bot,\] we have \[\operatorname{proj}_W (ku) = ku_1 = k\operatorname{proj}_W u.\] Therefore \(\operatorname{proj}_W\) is a linear transformation.

Next we find the image of \(\operatorname{proj}_W.\) For any \(v\in V,\) \(\operatorname{proj}_W v\in W,\) and we have \(\operatorname{Im}(\operatorname{proj}_W)\subseteq W.\) Conversely, for any \(w\in W,\) take \(v=w\) then \(\operatorname{proj}_W v = w,\) and we have \(W\subseteq \operatorname{Im}(\operatorname{proj}_W ).\) Hence \(\operatorname{Im}(\operatorname{proj}_W) = W.\)

In the same manner, it is easily shown that \(\operatorname{Ker}(\operatorname{proj}_W) = W^\bot.\) (Use a basis for \(W.\))

Solution. \[\operatorname{proj}_W \circ \operatorname{proj}_W = \operatorname{proj}_W.\]

Solution. Let \(\dim(V) = n\) and \(\dim(W)=m.\) Since \(W\) is finite-dimensional, we can construct an orthonormal basis for \(W\): \[v_1 ,\, v_2 ,\, \cdots ,\, v_m.\] Since \(V\) is finite-dimensional, we can extend this family to an orthonormal basis \(B\) for \(V\): \[v_1 ,\, v_2 ,\, \cdots ,\, v_m ,\, \cdots ,\, v_n.\] With this basis, we have \[M_{B,B}(\operatorname{proj}_W) = \left( \begin{array}{c|c} I_m & 0 \\ \hline 0 & 0 \end{array} \right).\]

Solution. Let \(\phi : \mathbb{C} \rightarrow \mathbb{C}\) be given by \[\phi(z) = \bar{z}\] for \(z\in \mathbb{C}.\) For any \(z\in \mathbb{C},\) we have \[\phi(\bar{z}) = \bar{\bar{z}} = z;\] Hence \(\phi\) is onto.

Solution. Define \(\phi : \mathbb{C}^{n} \rightarrow \mathbb{R}^{2n}\) by \[\phi(z_1 ,\, z_2 ,\, \cdots ,\,z_n ) = (\operatorname{Re}(z_1),\, \operatorname{Im}(z_1) ,\, \operatorname{Re}(z_2),\,\operatorname{Im}(z_2),\,\cdots,\,\operatorname{Re}(z_n),\,\operatorname{Im}(z_n)).\] Keep calm and show that \(\phi\) is an isomorphism.

Solution. \[\begin{aligned} \langle (2+5i ,\, 1-4i) \,\vert\, (2+5i ,\,1-4i)\rangle &= (2+5i)\overline{(2+5i)} + (1-4i)\overline{(1-4i)} \\[4pt] &= (2+5i)(2-5i) + (1-4i)(1+4i) \\[4pt] &= 4 + 25 + 1 + 16 = 46 , \\[4pt] \lVert (2+5i ,\, 1-4i ) \rVert &= \sqrt{46}. \end{aligned} \]

Solution. Recall that \[f(x) = e^{ix} = \cos x + i\sin x\] for \(x\in\mathbb{R}.\) Hence we have \[\begin{aligned} \int_{-\pi}^{\pi} \langle f\,\vert\,f \rangle dx &= \int_{-\pi}^{\pi} (\cos x + i \sin x)(\cos x - i\sin x) dx \\[4pt] &= \int_{-\pi}^{\pi} (\cos^2 x + \sin^2 x) dx \\[4pt] &= \int_{-\pi}^{\pi} 1\,dx = 2\pi ,\\[4pt] \lVert f \rVert &= \sqrt{2\pi}. \end{aligned}\]

Solution. If the inner product were defined without conjugation, a length of a vector might be complex. In this case, we cannot compare the distances of vectors.

Solution. \[\begin{aligned} \lVert av \rVert &= \sqrt{\langle av \,\vert\, av \rangle} \\[4pt] &= \sqrt{a\langle v\,\vert\, av \rangle} \\[4pt] &= \sqrt{a\overline{a} \langle v\,\vert\, v \rangle} \\[4pt] &= \sqrt{\lvert a \rvert^2 \langle v\,\vert\, v \rangle} \\[4pt] &= \lvert a \rvert \sqrt{ \langle v\,\vert\, v \rangle} \\[4pt] &= \lvert a \rvert \cdot \lVert v \rVert. \end{aligned}\]

Solution. For any \(v\in V,\) \(v\) is expressed in the unique form \[v = \lambda_1 v_1 + \lambda_2 v_2 + \cdots\] where \(\lambda_j\)s are complex numbers, and all but finite \(\lambda_j\)s are zero. Each \(\lambda_j\) is expressed in the unique form \[\lambda_j = a_j + ib_j\] where \(a_j\) and \(b_j\) are real numbers. Hence \[\begin{aligned} v &= (a_1 + ib_1)v_1 + (a_2 + ib_2)v_2 + \cdots \\[4pt] &= a_1 v_1 + b_1 iv_1 + a_2 v_2 + b_2 iv_2 + \cdots . \end{aligned}\] This expression is unique; Hence \(\left\{v_j ,\, iv_j \,\vert\,j\in J\right\}\) constitutes a basis for \(V\) over \(\mathbb{R}.\)

Solution. Let \(\theta\) be the angle between \(v\) and \(iv.\) \[\begin{gather} \langle v\,\vert\, iv \rangle = -i \langle v\,\vert\, v \rangle = -i \lVert v \rVert^2 \ne 0,\\[4pt] \cos\theta = \frac{\operatorname{Re}(\langle v\,\vert\, iv\rangle)}{\lVert v \rVert \lVert iv \rVert} = \frac{0}{\lVert v \rVert \lVert iv \rVert} =0. \end{gather}\]

Solution. Let \(z=a+bi,\) \(w=c+di\) where \(a,\) \(b,\) \(c,\) \(d\) are real numbers. First, observe that \[z\overline{w} = (a+bi)(c-di) = (ac+bd) + (bc-ad)i,\] hence \(z\overline{w}\) is purely imaginary if and only if \(ac+bd=0.\)

Next, in \(\mathbb{R}^2,\) since \([z]=(a,\,b)\) and \([w]=(c,\,d),\) we have \[\langle z\,\vert\,w\rangle = \langle (a,\,b) \,\vert\,(c,\,d)\rangle = ac+bd.\] Hence \(z\bot w\) if and only if \(ac+bd=0.\)

Therefore \(z\bot w\) in \(\mathbb{R}^2\) if and only if \(z\overline{w}\) is purely imaginary.

We now generalize this to higher dimensions. We only consider for the finite dimensional spaces. Let \(z,\,w\in \mathbb{C}^n,\) then \(z\) and \(w\) are expressible by \[\begin{aligned} z &= (z_1 ,\, z_2 ,\, \cdots ,\, z_n ),\\[4pt] w &= (w_1 ,\, w_2 ,\, \cdots ,\, w_n ),\\[4pt] z_j &= a_j + ib_j ,\\[4pt] w_j &= c_j + id_j \end{aligned}\] where \(a_j,\) \(b_j,\) \(c_j,\) \(d_j\) are real numbers.

First, observe that \[\begin{aligned} \langle z\,\vert\, w\rangle &= z_1 \overline{w_1} + \cdots + z_n \overline{w_n} \\[4pt] &= (a_1 + ib_1)(c_1 - id_1) + \cdots + (a_n +ib_n)(c_n-id_n) \\[4pt] &= [ (a_1 c_1 + b_1 d_1) + \cdots + (a_n c_n + b_n d_n )] + [ (b_1 c_1 - a_1 d_1 ) + \cdots + (b_n c_n -a_n d_n)]i. \end{aligned}\] Hence \(\langle z\,\vert\,w\rangle\) is purely imaginary if and only if \[(a_1 c_1 + b_1 d_1) + \cdots + (a_n c_n + b_n d_n )=0.\] Next, in \(\mathbb{R}^{2n},\) let \[\begin{aligned} \mathbf{x} &= (a_1 ,\, b_1 ,\, a_2 ,\, b_2 ,\, \cdots ,\, a_n ,\, b_n ),\\[4pt] \mathbf{y} &= (c_1 ,\, d_1 ,\, c_2 ,\, d_2 ,\, \cdots ,\, c_n ,\, d_n ). \end{aligned}\] Then \[\mathbf{x}\cdot\mathbf{y} = (a_1 c_1 + b_ 1 d_1) + \cdots + (a_n c_n + b_n d_n ).\] Hence \(\mathbf{x} \bot \mathbf{y}\) if and only if \[(a_1 c_1 + b_1 d_1) + \cdots + (a_n c_n + b_n d_n )=0.\] But for \([z] = \mathbf{x} ,\) \([w] = \mathbf{y}\) in \(\mathbb{R}^{2n},\) we conclude that \(\langle z\,\vert\,w\rangle\) is purely imaginary if and only if \(z\bot w\) in \(\mathbb{R}^{2n},\) that is, \(z\bot w\) in \(\mathbb{C}^n\) if and only if \(z\bot w\) in \(\mathbb{R}^{2n}.\)