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Chapter 7: Problem 49

Find the Fourier coefficient (component) \(c\) and the projection \(c w\) of \(v=(3+4 i, 2-3 i)\) along \(w=(5+i, 2 i)\) in \(\mathbf{C}^{2}\).

Short Answer

Expert verified
The Fourier coefficient \(c\) is \(\frac{7+22i}{29}\), and the projection \(cw\) is \((-\frac{7}{29}+\frac{71i}{29}, -\frac{30}{29}-\frac{14i}{29})\).

Step by step solution

01

Find the Inner Product between \(v\) and \(w\)

First, compute the inner product between \(v\) and \(w\). In this case, we are working in \(\mathbf{C}^{2}\), so we should use the complex inner product: \(\langle v, w \rangle = \sum_{k=1}^{2} v_k \overline{w_k}\) \(v = (3+4i, 2-3i)\) \(w = (5+i, 2i)\) \(\langle v, w \rangle = (3+4i)(5-i) + (2-3i)(-2i) = 15+12i-20i-4i^2+6i+3i^2 = 15+22i-8= 7 + 22i\)
02

Find the Normalization of \(w\)

Next, find the square of the normalization of \(w\) by computing the inner product of \(w\) with itself: \(\lVert w \rVert^2 = \langle w, w \rangle = (5+i)(5-i) + (2i)(-2i) = 25-5i+5i+i^2+4=29\)
03

Compute the Fourier Coefficient

Calculate the Fourier coefficient by taking the ratio of the inner product of \(v\) and \(w\) to the normalization of \(w\): \(c = \frac{\langle v, w \rangle}{\lVert w \rVert^2} = \frac{7+22i}{29}\)
04

Compute the Projection

Finally, find the projection \(cw\) by multiplying \(w\) by the Fourier coefficient found in Step 3: \(cw = (\frac{7+22i}{29})(5+i, 2i) = \frac{7+22i}{29}(5+i, 2i) =(\frac{35+29i+22i+22i^2}{29}, \frac{14i+44i^2}{29}) =\) \(=(-\frac{7}{29}+\frac{71i}{29}, -\frac{30}{29}-\frac{14i}{29})\) The Fourier coefficient \(c\) is \(\frac{7+22i}{29}\), and the projection \(cw\) is \((-\frac{7}{29}+\frac{71i}{29}, -\frac{30}{29}-\frac{14i}{29})\).

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Most popular questions from this chapter

Suppose \(P\) is an orthogonal matrix. Show that (a) \(\langle P u, P v\rangle=\langle u, v\rangle\) for any \(u, v \in V\) (b) \(\|P u\|=\|u\|\) for every \(u \in V\) Use \(P^{T} P=I\) and \(\langle u, v\rangle=u^{T} v\) (a) \(\langle P u, P v\rangle=(P u)^{T}(P v)=u^{T} P^{T} P v=u^{T} v=\langle u, v\rangle\) (b) We have \\[\|P u\|^{2}=\langle P u, P u\rangle=u^{T} P^{T} P u=u^{T} u=\langle u, u\rangle=\|u\|^{2}\\] Taking the square root of both sides gives our result.

Let \(W\) be the subspace of \(\mathbf{R}^{5}\) spanned by \(u=(1,2,3,-1,2)\) and \(v=(2,4,7,2,-1) .\) Find a basis of the orthogonal complement \(W^{\perp}\) of \(W\) We seck all vectors \(w=(x, y, z, s, t)\) such that \\[\begin{array}{l}\langle w, u\rangle=x+2 y+3 z-s+2 t=0 \\ \langle w, v\rangle=2 x+4 y+7 z+2 s-t=0\end{array}\\] Eliminating \(x\) from the second equation, we find the equivalent system \\[\begin{array}{r}x+2 y+3 z-s+2 t=0 \\\z+4 s-5 t=0\end{array}\\] The free variables are \(y, s,\) and \(t .\) Therefore, (1) \(\operatorname{Set} y=-1, s=0, t=0\) to obtain the solution \(w_{1}=(2,-1,0,0,0)\) (2) \(\operatorname{Set} y=0, s=1, t=0\) to find the solution \(w_{2}=(13,0,-4,1,0)\) (3) \(\operatorname{Set} y=0, s=0, t=1\) to obtain the solution \(w_{3}=(-17,0,5,0,1)\) The set \(\left\\{w_{1}, w_{2}, w_{3}\right\\}\) is a basis of \(W^{\perp}\)

Prove Theorem 7.17: Let \(A\) be the matrix representation of any inner product on \(V\). Then \(A\) is a positive definite matrix. Because \(\left\langle w_{i}, w_{j}\right\rangle=\left\langle w_{j}, w_{i}\right\rangle\) for any basis vectors \(w_{i}\) and \(w_{j},\) the matrix \(A\) is symmetric. Let \(X\) be any nonzero vector in \(\mathbf{R}^{n}\). Then \([u]=X\) for some nonzero vector \(u \in V\). Theorem 7.16 tells us that \(X^{T} A X=[u]^{T} A[u]=\langle u, u\rangle>0 .\) Thus, \(A\) is positive definite.

Prove Theorem 7.8: Suppose \(w_{1}, w_{2}, \ldots, w_{r}\) form an orthogonal set of nonzero vectors in \(V .\) Let \(v \in V\). Define \\[v^{\prime}=v-\left(c_{1} w_{1}+c_{2} w_{2}+\cdots+c_{r} w_{r}\right), \quad \text { where } \quad c_{i}=\frac{\left\langle v, w_{i}\right\rangle}{\left\langle w_{i}, w_{i}\right\rangle}\\] Then \(v^{\prime}\) is orthogonal to \(w_{1}, w_{2}, \dots, w_{r}\) For \(i=1,2, \ldots, r\) and using \(\left\langle w_{i}, w_{j}\right\rangle=0\) for \(i \neq j,\) we have \\[\begin{aligned} \left\langle v-c_{1} w_{1}-c_{2} x_{2}-\cdots-c_{r} w_{r}, w_{i}\right\rangle &=\left\langle v, w_{i}\right\rangle-c_{1}\left\langle w_{1}, w_{i}\right\rangle-\cdots-c_{i}\left\langle w_{i}, w_{i}\right\rangle-\cdots- c_{r}\left\langle w_{r}, w_{i}\right\rangle \\ &=\left\langle v, w_{i}\right\rangle-c_{1} \cdot 0-\cdots-c_{i}\left\langle w_{i}, w_{i}\right\rangle-\cdots-c_{r} \cdot 0 \\ &=\left\langle v, w_{i}\right\rangle-c_{i}\left\langle w_{i}, w_{i}\right\rangle=\left\langle v, w_{i}\right\rangle-\frac{\left\langle v, w_{i}\right\rangle}{\left\langle w_{i}, w_{i}\right\rangle}\left\langle w_{i}, w_{i}\right\rangle=0 \end{aligned}\\] The theorem is proved.

Prove Theorem 7.1 (Cauchy-Schwarz): For \(u\) and \(v\) in a real inner product space \(V\) \(\langle u, u\rangle^{2} \leq\langle u, u\rangle\langle v, v\rangle \quad\) or \(\quad|\langle u, v\rangle| \leq\|u\|\|v\|\) For any real number \(t\) \\[\langle t u+v, \quad t u+v\rangle=t^{2}\langle u, u\rangle+2 t(u, v)+\langle v, v\rangle=t^{2}\|u\|^{2}+2 t(u, v\rangle+\|v\|^{2}\\] Let \(a=\|u\|^{2}, b=2\langle u, v), c=\|v\|^{2} .\) Because \(\|t u+v\|^{2} \geq 0,\) we have \\[a t^{2}+b t+c \geq 0\\] for every value of \(t .\) This means that the quadratic polynomial cannot have two real roots, which implies that \(b^{2}-4 a c \leq 0\) or \(b^{2} \leq 4 a c .\) Thus, \\[4\langle u, v\rangle^{2} \leq 4\|u\|^{2}\|v\|^{2}\\] Dividing by 4 gives our result.
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