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Chapter 7: Problem 53

Find the matrix \(P\) that represents the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\\{1, i, 1-i\\}\).

Short Answer

Expert verified
The matrix \(P\) representing the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\{1, i, 1-i\}\) is: \( P =\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 2 \end{pmatrix}\).

Step by step solution

01

Express the basis vectors in terms of the standard basis

Given the basis \(\{1, i, 1-i\}\), we want to express each basis vector as a column vector with respect to the standard basis \(\{e_1, e_2, e_3\}\): \(e_1=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}\), \(e_2=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}\), and \(e_3=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}\). Our basis vectors can then be written as: \(v_1=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}\), \(v_2=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix}\), and \(v_3=\begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}\).
02

Compute the inner products of the basis vectors

Now, we calculate the inner products of the basis vectors: \(\langle v_1, v_1 \rangle = \langle e_1, e_1 \rangle = 1, \\ \langle v_1, v_2 \rangle= \langle v_2, v_1 \rangle=\langle e_1, e_2 \rangle = 0, \\ \langle v_1, v_3 \rangle= \langle v_3, v_1 \rangle=\langle e_1, e_3 \rangle = 1, \\ \langle v_2, v_2 \rangle=\langle e_2, e_2 \rangle = 1, \\ \langle v_2, v_3 \rangle= \langle v_3, v_2 \rangle= \langle e_2, e_3 \rangle = 0, \\ \langle v_3, v_3 \rangle=\langle e_3, e_3 \rangle = 2\).
03

Construct the matrix \(P\)

Using the above results, we can construct the matrix \(P\): \(P = \begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 2 \end{pmatrix}\). Now, to compute the inner product of any two vectors \(x\) and \(y\) with respect to the given basis, we can use the formula: \(\langle x,y \rangle = x^{*}Py\). The matrix \(P\) representing the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\{1, i, 1-i\}\) is: \( P =\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 2 \end{pmatrix}\).

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Most popular questions from this chapter

Prove Theorem \(7.14: A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]\) is positive definite if and only if \(a\) and \(d\) are positive and Let \(u=[x, y]^{T} .\) Then \\[f(u)=u^{T} A u=[x, y]\left[\begin{array}{ll} a & b \\ b & d\end{array}\right]\left[\begin{array}{l} x \\\y \end{array}\right]=a x^{2}+2 b x y+d y^{2}\\] Suppose \(f(u)>0\) for every \(u \neq 0 .\) Then \(f(1,0)=a>0\) and \(f(0,1)=d>0 .\) Also, we have \(f(b,-a)=a\left(a d-b^{2}\right)>0 .\) Because \(a>0,\) we get \(a d-b^{2}>0\) Conversely, suppose \(a>0, b=0, a d-b^{2}>0 .\) Completing the square gives us \\[f(u)=a\left(x^{2}+\frac{2 b}{a} x y+\frac{b^{2}}{a_{2}} y^{2}\right)+d y^{2}-\frac{b^{2}}{a} y^{2}=a\left(x+\frac{b y}{a}\right)^{2}+\frac{a d-b^{2}}{a} y^{2}\\] Accordingly, \(f(u)>0\) for every \(u \neq 0\)

Suppose \(S, S_{1}, S_{2}\) are the subsets of \(V\). Prove the following: (a) \(S \subseteq S^{\perp \perp}\) (b) If \(S_{1} \subseteq S_{2},\) then \(S_{2}^{\perp} \subseteq S_{1}^{\perp}\) (c) \(S^{\perp}=\operatorname{span}(S)^{\perp}\) (a) Let \(w \in S\). Then \(\langle w, v\rangle=0\) for every \(v \in S^{\perp} ;\) hence, \(w \in S^{\perp \perp}\). Accordingly, \(S \subseteq S^{\perp \perp}\). (b) Let \(w \in S_{2}^{\perp} .\) Then \(\langle w, v\rangle=0\) for every \(v \in S_{2} .\) Because \(S_{1} \subseteq S_{2},\langle w, v\rangle=0\) for every \(v=S_{1} .\) Thus, \(w \in S_{1}^{\perp},\) and hence, \(S_{2}^{\perp} \subseteq S_{1}^{\perp}\) (c) Because \(S \subseteq \operatorname{span}(S),\) part (b) gives us \(\operatorname{span}(S)^{\perp} \subseteq S^{\perp} .\) Suppose \(u \in S^{\perp}\) and \(v \in \operatorname{span}(S) .\) Then there exist \(w_{1}, w_{2}, \ldots, w_{k}\) in \(S\) such that \(v=a_{1} w_{1}+a_{2} w_{2}+\dots+a_{k} w_{k} .\) Then, using \(u \in S^{\perp},\) we have \\[ \begin{aligned} \langle u, v\rangle &=\left\langle u, a_{1} w_{1}+a_{2} w_{2}+\cdots+a_{k} w_{k}\right\rangle=a_{1}\left\langle u, w_{1}\right\rangle+a_{2}\left\langle u, w_{2}\right\rangle+\cdots+a_{k}\left\langle u, w_{k}\right\rangle \\ &=a_{1}(0)+a_{2}(0)+\cdots+a_{k}(0)=0\end{aligned}\\] Thus, \(u \in \operatorname{span}(S)^{\perp} .\) Accordingly, \(S^{\perp} \subseteq \operatorname{span}(S)^{\perp} .\) Both inclusions give \(S^{\perp}=\operatorname{span}(S)^{\perp}\)

Prove Theorem 7.7: Let \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) be an orthogonal basis of \(V .\) Then for any \(v \in V\) \\[v=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle} u_{1}+\frac{\left\langle v, u_{2}\right\rangle}{\left\langle u_{2}, u_{2}\right\rangle} u_{2}+\cdots+\frac{\left\langle v, u_{n}\right\rangle}{\left\langle u_{n}, u_{n}\right\rangle} u_{n}\\] Suppose \(v=k_{1} u_{1}+k_{2} u_{2}+\cdots+k_{n} u_{n} .\) Taking the inner product of both sides with \(u_{1}\) yields \\[\begin{aligned} \left\langle v, u_{1}\right\rangle &=\left\langle k_{1} u_{2}+k_{2} u_{2}+\cdots+k_{n} u_{n}, u_{1}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{1}\right\rangle+k_{2}\left\langle u_{2}, u_{1}\right\rangle+\cdots+k_{n}\left\langle u_{n}, u_{1}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{1}\right\rangle+k_{2} \cdot 0+\cdots+k_{n} \cdot 0=k_{1}\left\langle u_{1}, u_{1}\right\rangle \end{aligned} \\] Thus, \(k_{1}=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle} .\) Similarly, for \(i=2, \ldots, n\) \\[ \begin{aligned} \left\langle v, u_{i}\right\rangle &=\left\langle k_{1} u_{i}+k_{2} u_{2}+\cdots+k_{n} u_{n}, u_{i}\right\rangle \\ &=k_{1}\left\langle u_{1}, u_{i}\right\rangle+k_{2}\left\langle u_{2}, u_{i}\right\rangle+\cdots+k_{n}\left\langle u_{n}, u_{i}\right\rangle \\ &=k_{1} \cdot 0+\cdots+k_{i}\left\langle u_{i}, u_{i}\right\rangle+\cdots+k_{n} \cdot 0=k_{i}\left\langle u_{i}, u_{i}\right\rangle\end{aligned}\\] Thus, \(k_{i}=\frac{\left\langle v, u_{i}\right\rangle}{\left\langle u_{1}, u_{i}\right\rangle} .\) Substituting for \(k_{i}\) in the equation \(v=k_{1} u_{1}+\cdots+k_{n} u_{n},\) we obtain the desired result.

Verify that the following is an inner product on \(\mathbf{R}^{2}\), where \(u=\left(x_{1}, x_{2}\right)\) and \(v=\left(y_{1}, y_{2}\right)\)

Find an orthogonal matrix \(P\) whose first row is \(u_{1}=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)\) First find a nonzero vector \(w_{2}=(x, y, z)\) that is orthogonal to \(u_{1}-\) that is, for which \\[0=\left\langle u_{1}, w_{2}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0\\] One such solution is \(w_{2}=(0,1,-1) .\) Normalize \(w_{2}\) to obtain the second row of \(P\) \\[u_{2}=(0,1 / \sqrt{2},-1 / \sqrt{2})\\] Next find a nonzero vector \(w_{3}=(x, y, z)\) that is orthogonal to both \(u_{1}\) and \(u_{2}\) -that is, for which \\[\begin{array}{l} 0=\left\langle u_{1}, w_{3}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0 \\ 0=\left\langle u_{2}, w_{3}\right\rangle=\frac{y}{\sqrt{2}}-\frac{y}{\sqrt{2}}=0 \quad \text { or } \quad y-z=0 \end{array}\\] Set \(z=-1\) and find the solution \(w_{3}=(4,-1,-1) .\) Normalize \(w_{3}\) and obtain the third row of \(P ;\) that is, \\[u_{3}=(4 / \sqrt{18},-1 / \sqrt{18},-1 / \sqrt{18})\\] Thus, \\[P=\left[\begin{array}{ccc}\frac{1}{5} & \frac{2}{3} & \frac{2}{3} \\ 0 & 1 / \sqrt{2} & -1 / \sqrt{2} \\ 4 / 3 \sqrt{2} & -1 / 3 \sqrt{2} & -1 / 3 \sqrt{2} \end{array}\right]\\] We emphasize that the above matrix \(P\) is not unique.
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