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Chapter 7: Problem 98

Find the matrix \(P\) that represents the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\\{1,1+i, 1-2 i\\}\).

Short Answer

Expert verified
The matrix P that represents the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\{1, 1+i, 1-2 i\}\) is: \[ P = \begin{bmatrix} 1 & 1 - i & 1 + 2i \\ 1 + i & 2 & 3 - 5i \\ 1 - 2i & 3 + 5i & 5 \end{bmatrix}\]

Step by step solution

01

Calculate the inner products of the basis vectors.

Let's denote the basis vectors as follows: \(v_1 = 1\) \(v_2 = 1 + i\) \(v_3 = 1 - 2i\) We now need to calculate the inner products of each pair of basis vectors: \(\langle v_1, v_1 \rangle,\; \langle v_1, v_2 \rangle,\; \langle v_1, v_3 \rangle,\; \langle v_2, v_1 \rangle,\; \langle v_2, v_2 \rangle,\; \langle v_2, v_3 \rangle,\; \langle v_3, v_1 \rangle,\; \langle v_3, v_2 \rangle,\; and\; \langle v_3, v_3 \rangle\) We use the definition of the inner product: \(\langle u, v \rangle = u_1 \overline{v_1} + u_2 \overline{v_2} + u_3 \overline{v_3}\) for \(u = (u_1, u_2, u_3)\) and \(v = (v_1, v_2, v_3)\). Calculating the inner products: \(\langle v_1, v_1 \rangle = (1)(\overline{1}) + (0)(\overline{0}) + (0)(\overline{0}) = 1\) \(\langle v_1, v_2 \rangle = (1)(\overline{(1 + i)}) + (0)(\overline{0}) + (0)(\overline{0}) = 1 - i\) \(\langle v_1, v_3 \rangle = (1)(\overline{(1 - 2i)}) + (0)(\overline{0}) + (0)(\overline{0}) = 1 + 2i\) \(\langle v_2, v_1 \rangle = \langle v_1, v_2 \rangle^* = (1 - i)^*(=) 1 + i\) \(\langle v_2, v_2 \rangle = (1 + i)(\overline{(1 + i)})+(0)(\overline{0}) + (0)(\overline{0}) = 2\) \(\langle v_2, v_3 \rangle = (1 + i)(\overline{(1 - 2i)}) + (0)(\overline{0}) + (0)(\overline{0}) = 3 - 5i\) \(\langle v_3, v_1 \rangle = \langle v_1, v_3 \rangle^* = (1 + 2i)^* = 1 - 2i\) \(\langle v_3, v_2 \rangle = \langle v_2, v_3 \rangle^* = (3 - 5i)^* = 3 + 5i\) \(\langle v_3, v_3 \rangle = (1 - 2i)(\overline{(1 - 2i)}) + (0)(\overline{0}) + (0)(\overline{0}) = 5\)
02

Form the Hermitian matrix P using the calculated inner products.

Now we will create the matrix P by using the calculated inner products: \[ P = \begin{bmatrix} \langle v_1, v_1 \rangle & \langle v_1, v_2 \rangle & \langle v_1, v_3 \rangle \\ \langle v_2, v_1 \rangle & \langle v_2, v_2 \rangle & \langle v_2, v_3 \rangle \\ \langle v_3, v_1 \rangle & \langle v_3, v_2 \rangle & \langle v_3, v_3 \rangle \end{bmatrix}\] Inserting the calculated inner product values: \[ P = \begin{bmatrix} 1 & 1 - i & 1 + 2i \\ 1 + i & 2 & 3 - 5i \\ 1 - 2i & 3 + 5i & 5 \end{bmatrix}\] Therefore, the matrix P that represents the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\{1, 1+i, 1-2 i\}\) is: \[ P = \begin{bmatrix} 1 & 1 - i & 1 + 2i \\ 1 + i & 2 & 3 - 5i \\ 1 - 2i & 3 + 5i & 5 \end{bmatrix}\]

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Most popular questions from this chapter

Verify each of the following: (a) Parallelogram Law (Fig. 7-7): \(\|u+v\|^{2}+\|u-v\|^{2}=2\|u\|^{2}+2\|v\|^{2}\) (b) Polar form for \(\langle u, v\rangle\) (which shows the inner product can be obtained from the norm function): \\[ \langle u, v\rangle=\frac{1}{4}\left(\|u+v\|^{2}-\|u-v\|^{2}\right) \\] Expand as follows to obtain \\[\begin{array}{l}\|u+v\|^{2}=\langle u+v, u+v\rangle=\|u\|^{2}+2\langle u, v\rangle+\|v\|^{2} \\ \|u-v\|^{2}=\langle u-v, u-v\rangle=\|u\|^{2}-2\langle u, v\rangle+\|v\|^{2}\end{array}\\] Add (1) and (2) to get the Parallelogram Law (a). Subtract (2) from (1) to obtain \\[\|u+v\|^{2}-\|u-v\|^{2}=4\langle u, v\rangle\\] Divide by 4 to obtain the (real) polar form (b).

Suppose \(E=\left\\{e_{1}, e_{2}, \ldots, e_{n}\right\\}\) is an orthonormal basis of \(V .\) Prove (a) For any \(u \in V,\) we have \(u=\left\langle u, e_{1}\right\rangle e_{1}+\left\langle u, e_{2}\right\rangle e_{2}+\cdots+\left\langle u, e_{n}\right\rangle e_{n}\) (b) \(\left\langle a_{1} e_{1}+\cdots+a_{n} e_{n}, \quad b_{1} e_{1}+\cdots+b_{n} e_{n}\right\rangle=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\) (c) For any \(u, v \in V,\) we have \(\langle u, v\rangle=\left\langle u, e_{1}\right\rangle\left\langle v, e_{1}\right\rangle+\cdots+\left\langle u, e_{n}\right\rangle\left\langle v, e_{n}\right\rangle\) (a) Suppose \(u=k_{1} e_{1}+k_{2} e_{2}+\cdots+k_{n} e_{n} .\) Taking the inner product of \(u\) with \(e_{1}\) \\[\begin{aligned}\left\langle u, e_{1}\right\rangle &=\left\langle k_{1} e_{1}+k_{2} e_{2}+\cdots+k_{n} e_{n}, e_{1}\right\rangle \\ &=k_{1}\left\langle e_{1}, e_{1}\right\rangle+k_{2}\left\langle e_{2}, e_{1}\right\rangle+\cdots+k_{n}\left\langle e_{n}, e_{1}\right\rangle \\ &=k_{1}(1)+k_{2}(0)+\cdots+k_{n}(0)=k_{1}\end{aligned}\\] Similarly, for \(i=2, \ldots, n\) \\[\begin{aligned}\left\langle u, e_{i}\right\rangle &=\left\langle k_{1} e_{1}+\cdots+k_{i} e_{i}+\cdots+k_{n} e_{n}, \quad e_{i}\right\rangle \\\&=k_{1}\left\langle e_{1}, e_{i}\right\rangle+\cdots+k_{i}\left\langle e_{i}, e_{i}\right\rangle+\cdots+k_{n}\left\langle e_{n}, e_{i}\right\rangle \\\ &=k_{1}(0)+\cdots+k_{i}(1)+\cdots+k_{n}(0)=k_{i} \end{aligned}\\] Substituting \(\left\langle u, e_{i}\right\rangle\) for \(k_{i}\) in the equation \(u=k_{1} e_{1}+\cdots+k_{n} e_{n},\) we obtain the desired result. (b) We have \\[\left\langle\sum_{i=1}^{n} a_{i} e_{i}, \sum_{j=1}^{n} b_{j} e_{j}\right\rangle=\sum_{i, j=1}^{n} a_{i} b_{j}\left\langle e_{i}, e_{j}\right\rangle=\sum_{i=1}^{n} a_{i} b_{i}\left\langle e_{i}, e_{i}\right\rangle+\sum_{i \neq j} a_{i} b_{j}\left\langle e_{i}, e_{j}\right\rangle\\] But \(\left\langle e_{i}, e_{j}\right\rangle=0\) for \(i \neq j,\) and \(\left\langle e_{i}, e_{j}\right\rangle=1\) for \(i=j .\) Hence, as required, \\[\left\langle\sum_{i=1}^{n} a_{i} e_{i}, \sum_{j=1}^{n} b_{j} e_{j}\right\rangle=\sum_{i=1}^{n} a_{i} b_{i}=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\\] (c) By part (a), we have \\[u=\left\langle u, e_{1}\right\rangle e_{1}+\cdots+\left\langle u, e_{n}\right\rangle e_{n} \quad \text { and } \quad v=\left\langle v,e_{1}\right\rangle e_{1}+\cdots+\left\langle v, e_{n}\right\rangle e_{n}\\] Thus, by part (b), \\[\langle u, v\rangle=\left\langle u, e_{1}\right\rangle\left\langle v, e_{1}\right\rangle+\left\langle u, e_{2}\right\rangle\left\langle v, e_{2}\right\rangle+\cdots+\left\langle u, e_{n}\right\rangle\left\langle v, e_{n}\right\rangle\\]

Which of the following symmetric matrices are positive definite? (a) \(A=\left[\begin{array}{ll}3 & 4 \\ 4 & 5\end{array}\right]\) (b) \(B=\left[\begin{array}{rr}8 & -3 \\ -3 & 2\end{array}\right]\) (c) \(C=\left[\begin{array}{rr}2 & 1 \\ 1 & -3\end{array}\right]\) (d) \(D=\left[\begin{array}{ll}3 & 5 \\ 5 & 9\end{array}\right]\) Use Theorem 7.14 that a \(2 \times 2\) real symmetric matrix is positive definite if and only if its diagonal entries are positive and if its determinant is positive. (a) No, because \(|A|=15-16=-1\) is negative. (b) Yes. (c) No, because the diagonal entry -3 is negative. (d) Yes.

Find an orthonormal basis of the subspace \(W\) of \(\mathbf{C}^{3}\) spanned by \\[v_{1}=(1, i, 0) \quad \text { and } \quad v_{2}=(1,2,1-i)\\] Apply the Gram-Schmidt algorithm. Set \(w_{1}=v_{1}=(1, i, 0) .\) Compute \\[v_{2}-\frac{\left\langle v_{2}, w_{1}\right\rangle}{\left\langle w_{1}, w_{1}\right\rangle} w_{1}=(1,2,1-i)-\frac{1-2 i}{2}(1, i, 0)=\left(\frac{1}{2}+i, 1-\frac{1}{2} i, 1-i\right)\\]

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