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Homework 9

  1. (25 p.) Consider the \(\mathbb{C}\)-vector space \(\mathbb{C}^{2\times 2}\) with the inner product given by \(\langle A,B \rangle = \text{tr}(B^*A)\). Find an orthonormal basis for the subspace of symmetric matrices \(S = \{A \in \mathbb{C}^{2\times 2} | A^t = A\}\).
    Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\), \(A^t = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \implies b=c\).
    Thus \(B_S = \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = v_1, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = v_2, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = v_3 \}\).
    Let \(z_1 = v_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), \(z_{2}=v_{2}-\frac{\langle v_{2},z_{1}\rangle}{||z_{1}||^{2}}z_{1}\)
    ​\(\langle v_2, z_1 \rangle = \text{tr}(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}^* \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}) = \text{tr}(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}) = \text{tr}(\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}) = 0\), then \(z_2 = v_2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).
    ​\(z_{3} = v_{3} - \frac{\langle v_{3}, z_{1} \rangle}{||z_{1}||^{2}}z_{1} - \frac{\langle v_{3}, z_{2} \rangle}{||z_{2}||^{2}}z_{2}\). And \(\langle v_{3}, z_{1} \rangle = \text{tr}( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}^{*} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}) = 0\).
    ​\(\langle v_{3}, z_{2} \rangle = \text{tr}( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}^{*} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}) = 0\).
    Thus orthogonal basis: \(\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \}\).
    And \(||v_1||^2 = \text{tr}(\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}) = 1\), \(||v_2||^2 = \text{tr}(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}) = 2\), \(||v_{3}||^{2}=\text{tr}( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix})=1\)

    Thus orthonormal basis: \(\left\{ \begin{pmatrix} 1 &0 \\ 0&0 \end{pmatrix}, \begin{pmatrix} 0& \frac{\sqrt2}{2} \\ \frac{\sqrt{2}}{2} &0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0&1 \end{pmatrix}\right\}\)​

  2. (25 p.) Let \(V\) be a vector space with inner product \(\langle -,-\rangle\). Suppose \(W \subseteq V\) is a finite-dimensional subspace of \(V\). Prove that if \(x \notin W\), then there exists \(y \in W^\perp\) such that \(\langle x, y \rangle \neq 0\).

    Proof

    We know \(V=W \oplus W^\perp\), and if \(x \notin W\), then \(x = w+w_\perp\), where \(w\) can be zero, and \(w_\perp \neq 0\), and \(w \in W\), \(w_\perp \in W^\perp\).

    Since \(y \in W^\perp\), then take \(y=w_\perp\). Then \(\langle x,y \rangle = \langle w+w_\perp, w_\perp \rangle = \langle w,w_\perp \rangle + \langle w_\perp, w_\perp \rangle = 0 + ||w_\perp||^2\). And since \(w_\perp \neq 0\), then \(\langle x,y \rangle > 0\). \(\implies \langle x,y \rangle \neq 0\).

  3. (25 p.) Consider the \(\mathbb{R}\)-vector space of continuous real-valued functions \(C[-1, 1]\) with the inner product given by \(\langle f, g \rangle = \int_{-1}^1 f(x) g(x) dx.\)
    Find the best approximation to the function \(f(x) = \sin(\pi x)\) by vectors in the subspace of polynomial functions \(\mathbb{R}_3[x]\).

    Proof

    First, find orthogonal basis of \(\mathbb{R}_3[x]\). \(B = \{1=v_1, x=v_2, x^2=v_3, x^3=v_4\}\).
    Let \(z_1=1\), \(z_2=v_2 - \frac{\langle v_2, z_1 \rangle}{\|z_1\|^2} z_1\). \(\langle v_2, z_1 \rangle = \int_{-1}^1 x dx = \frac{1}{2} - \frac{1}{2} = 0\).
    Then \(z_2 = v_2 = x\). \(z_3=v_3-\frac{\langle v_3,z_1\rangle}{\|z_1\|^2}z_1-\frac{\langle v_3,z_2\rangle}{\|z_2\|^2}z_2\). \(\langle v_3, z_1 \rangle = \int_{-1}^1 x^2 dx = \frac{2}{3}\).
    ​\(\|z_{1}\|^{2}=2\). \(\langle v_3, z_2 \rangle = \int_{-1}^1 x^3 dx = 0\). Thus, \(z_{3}=v_{3}-\frac{1}{3}z_{1}=x^{2}-\frac{1}{3}\).

    So, \(z_{4}=v_{4}-\frac{\langle v_{4},z_{1}\rangle}{\|z_{1}\|^{2}}z_{1}-\frac{\langle v_{4},z_{2}\rangle}{\|z_{2}\|^{2}}z_{2}-\frac{\langle v_{4},z_{3}\rangle}{\|z_{3}\|^{2}} z_{3}\). \(\langle v_4, z_1 \rangle = \int_{-1}^1 x^3 dx = 0\).

    \(\langle v_4, z_2 \rangle = \int_{-1}^1 x^4 dx = \frac{2}{5}\). \(\|z_2\|^2 = \int_{-1}^1 x^2 dx = \frac{2}{3}\).

    \(\langle v_{4}, z_{3}\rangle = \int_{-1}^{1}x^{3}(x^{2}- \frac{1}{3}) dx = \int_{-1} ^{1}(x^{5}- \frac{1}{3}x^{3}) dx = 0 - 0 = 0\).

    Thus \(z_4 = x^3 - \frac{3}{5}x\). Thus the orthogonal basis is: \(\{1,x,x^{2}-\frac{1}{3},x^{3}-\frac{3}{5}x\}\)
    Since \(||z_{1}||^{2}= 2, ||z_{2}||^{2}= \frac{2}{3}, ||z_{3}||^{2}= \frac{8}{45}, || z_{4} ||^{2}= \frac{8}{175}\)
    Then \(P_{s}(\sin(\pi x)) = \sum_{i=1}^{4}\frac{\lang\sin(\pi x), z_{i}\rang}{||z_{i}||^{2}} z_{i}\) \(=\frac{1}{2}\lang\sin(\pi x), 1\rang + \frac{\lang\sin(\pi x), x\rang}{\frac{2}{3}} x + \frac{\lang\sin(\pi x), x^{2}- \frac{1}{3}\rang}{\frac{8}{45}}(x^{2}- \frac{1}{3} ) + \frac{\lang\sin(\pi x), x^{3}- \frac{3}{5}x\rang}{\frac{8}{175}}(x^{3}- \frac{3}{5} x)\)

    \(\lang\sin(\pi x), 1\rang = \int_{-1}^{1}\sin(\pi x) dx = -\frac{\cos(\pi x)}{\pi} \Big|_{-1}^{1}= \frac{1}{\pi}- \frac{1}{\pi}= 0\)
    ​\(\lang\sin(\pi x), x\rang = \int_{-1}^{1} \sin(\pi x) \cdot x dx = -\frac{x}{\pi}\cos(\pi x)\Big|_{-1}^{1} - \int_{-1}^{1} \sin(\pi x) dx\) \(= \frac{2}{\pi}\)

    \(\left\langle\sin(\pi x),x^{2}-\frac{1}{3}\right\rangle=\int_{-1}^{1}\sin(\pi x)\left (x^{2}-\frac{1}{3}\right)dx=-\left.\frac{\cos(\pi x)}{\pi}\left(x^{2}-\frac{1}{3} \right)\right|_{-1}^{1}-\int_{-1}^{1}-\frac{1}{\pi}\cos(\pi x)\cdot2xdx\)

    \(=\frac{2}{\pi}\left[\left.\frac{x}{\pi}\sin\pi x\right|_{-1}^{1}-\int_{-1}^{1}\cos (\pi x)dx\right]=-\frac{2}{\pi}\int_{-1}^{1}\cos(\pi x)dx=-\frac{2}{\pi}(\sin(\pi )-\sin(-\pi))=0\)

    \(\left\langle\sin(\pi x),x^3-\frac35x\right\rangle=\int_{-1}^1\sin(\pi x)\left(x^3-\frac35x\right)dx=-\left.\frac{1}{\pi}\cos(\pi x)\left(x^3-\frac35x\right)\right|_{-1}^1-\int_{-1}^1-\frac{1}{\pi}\cos(\pi x)\left(3x^2-\frac35\right)dx\)

    \(=\frac{4}{5\pi}+\frac{1}{\pi}\int_{-1}^1\cos(\pi x)\left(3x^2-\frac35\right)dx=\frac{4}{5\pi}-\frac{1}{\pi}\times\frac{12}{\pi^2}=\frac{4}{5\pi}-\frac{12}{\pi^3}\)

    \(P_{s}(\sin(\pi x))=\frac{3}{\pi}x+\frac{35}{2\pi}x^{3}-\frac{525}{2\pi^{3}}x^{3} -\frac{21}{2\pi}x+\frac{315}{2\pi^{3}}x\)​

  4. (25 p.) Consider \(\mathbb{R}^4\) with the standard inner product, and let \(S_1\), \(S_2\), and \(S_3\) be the subspaces of \(\mathbb{R}^4\) defined by
    ​\(S_1: \begin{cases} x_1 + x_2 - 2x_3 = 0 \\ x_1 - x_2 = 0 \\ x_2 - x_4 = 0 \end{cases} \quad S_2: \begin{cases} x_1 + x_2 - x_3 + x_4 = 0 \\ 2x_1 + 2x_4 = 0 \end{cases} \quad S_3: \{2x_1 + x_2 + 2x_3 = 0\}\)
    Find an orthonormal basis \(\{v_1, v_2, v_3, v_4\}\) for \(\mathbb{R}^4\) such that \(v_i \in S_i\) (for \(i=1, 2, 3\)).

    \(S_1 = \text{span}((1,1,1,1))\), \(S_{2} = \text{span}((1,0,0,-1), (0,1,1,0))\), \(S_3 = \text{span}((1,-2,0,0), (0,-2,1,0), (0,0,0,1))\)

    Let \(v_1 = (1,1,1,1)\). \(v_{2} = a(1,0,0,-1) + b(0,1,1,0)\). \(v_3 = a'(1,-2,0,0) + b'(0,-2,1,0) + c'(0,0,0,1)\)
    ​\(\langle v_{2}, v_{1} \rangle = 0 = a+b+b-a=2b, \text{ then }v_{2}=(1,0,0,-1)\)
    ​\(\langle v_{3},v_{1}\rangle=0=a^{\prime}-2a^{\prime}-2b^{\prime}+b^{\prime}+c^{\prime} =-a^{\prime}-b^{\prime}+c^{\prime}\Rightarrow a^{\prime}=-b^{\prime}+c^{\prime}\)
    ​\(\langle v_3, v_2 \rangle = 0 = a'-c' \Rightarrow a'=c' \Rightarrow b'=0\)
    ​\(\text{Thus let } c'=1, a'=1, b'=0 \Rightarrow v_3=(1,-2,0,1)\)
    ​\(\text{Thus } v_1=(1,1,1,1), v_2=(1,0,0,-1), v_3=(1,-2,0,1).\)
    ​\(\text{Let } v_4=(a,b,c,d). \text{ Then } \langle v_1,v_4 \rangle = 0=a+b+c+d\)
    ​\(\langle v_2,v_4 \rangle = a-d=0, \langle v_3,v_4 \rangle = a-2b+d=0\)
    ​\(\Rightarrow \begin{cases} a+b+c+d=0 \\ a=d \\ a-2b+d=0 \end{cases} \Rightarrow \begin{cases} 2d+b+c=0 \\ a=d \\ b=d \end{cases} \Rightarrow c=-3d\)
    ​\(\text{Then let } d=1, v_4=(1,1,-3,1)\)

    Thus, \(\left\{ (1,1,1,1), (1,0,0,-1), (1,-2,0,1), (1,1,-3,1) \right\}\) is an orthogonal basis.
    \(||v_1||=2\), \(||v_2||=\sqrt{2}\), \(||v_3||=\sqrt{6}\), \(||v_4||=\sqrt{12}\)
    then \(\left\{ \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right), \left(\frac{\sqrt{2}}{2}, 0, 0, -\frac{\sqrt{2}}{2}\right), \left(\frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{3}, 0, \frac{\sqrt{6}}{6}\right), \left(\frac{\sqrt{3}}{6}, \frac{\sqrt{3}}{6}, -\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{6}\right) \right\}\) is an orthonormal basis such that \(v_i \in S_i\).

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