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In each case, determine whether the two vectors are orthogonal. a. \((4,-3 i, 2+i),(i, 2,2-4 i)\) b. \((i,-i, 2+i),(i, i, 2-i)\) c. \((1,1, i, i),(1, i,-i, 1)\) d. \((4+4 i, 2+i, 2 i),(-1+i, 2,3-2 i)\)

Short Answer

Expert verified
None of the vector pairs are orthogonal, as none of their dot products are zero.

Step by step solution

01

Understanding Orthogonality

Two vectors are orthogonal if their dot product is zero. The dot product of complex vectors is calculated using the formula: \( \mathbf{a} \cdot \mathbf{b} = a_1 \overline{b_1} + a_2 \overline{b_2} + a_3 \overline{b_3} + \dots \) where \( \overline{b_i} \) is the complex conjugate of \( b_i \). For a pair of vectors \( (a_{1}, a_{2}, a_{3}), (b_{1}, b_{2}, b_{3}) \), compute the dot product by substituting and simplifying.
02

Calculate Dot Product for Part (a)

Given vectors \((4, -3i, 2+i)\) and \((i, 2, 2-4i)\). Compute the dot product: \\[ 4 \cdot (-i) + (-3i) \cdot 2 + (2+i) \cdot (2+4i) = -4i + (-6i) + ((2+i)(2-4i)) \]. Now simplify the last term: \\[ (2+i)(2-4i) = 4 - 8i + 2i + 4 = 8 - 6i \]. The dot product becomes \\[ -4i - 6i + (8 - 6i) = 8 - 16i \]. Since this is not zero, the vectors are not orthogonal.
03

Calculate Dot Product for Part (b)

Given vectors \((i, -i, 2+i)\) and \((i, i, 2-i)\). Calculate the dot product: \\[ i \cdot (-i) + (-i) \cdot i + (2+i) \cdot (2-i) = 1 - 1 + ((2+i)(2-i)) \]. Simplify \((2+i)(2-i)\): \\[ (2+i)(2-i) = 4 - i^2 = 5 \]. So the dot product is \\[ 1 - 1 + 5 = 5 \]. As it is not zero, the vectors are not orthogonal.
04

Calculate Dot Product for Part (c)

Given vectors \((1, 1, i, i)\) and \((1, i, -i, 1)\). Compute the dot product: \\[ 1 \cdot 1 + 1 \cdot i + i \cdot (-i) + i \cdot 1 = 1 + i + 1 + i \]. Simplify to get \\[ 2 + 2i \]. Since this is not zero, the vectors are not orthogonal.
05

Calculate Dot Product for Part (d)

Given vectors \((4+4i, 2+i, 2i)\) and \((-1+i, 2, 3-2i)\). Calculate the dot product: \\[ (4+4i)(-1+i) + (2+i)(2) + (2i)(3-2i) \]. Calculate each term: \\[ (4+4i)(-1+i) = -4 - 4i + 4i + 4i^2 = -4 - 4(-1) = 0 \], \\[ (2+i)(2) = 4 + 2i \], and \\[ (2i)(3-2i) = 6i - 4i^2 = 6i + 4 = 4 + 6i \]. Add them: \\[ 0 + 4 + 2i + 4 + 6i = 8 + 8i \]. The result is not zero, so the vectors are not orthogonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product of Complex Vectors
The dot product is a core concept in understanding vector relationships, especially in complex vector spaces. In the realm of complex numbers, the traditional dot product formula undergoes a slight modification. Instead of just multiplying corresponding components of the vectors, we multiply by the complex conjugate of the second vector's component respectively. The formula used is:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1 \overline{b_1} + a_2 \overline{b_2} + a_3 \overline{b_3} + \dots \)
Here, \( \overline{b_i} \) denotes the complex conjugate of \( b_i \). This ensures that the product maintains real-valued properties analogous to real number vectors. The result from this dot product calculation determines if vectors in complex space are orthogonal or not.
Complex Conjugate
The concept of a complex conjugate is key when working with complex vectors. Simply put, the complex conjugate of a number is achieved by changing the sign of its imaginary part. For any complex number \( z = a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part, the complex conjugate is \( \overline{z} = a - bi \).Why is this important? When calculating the dot product of complex vectors, we use the complex conjugate to ensure the resulting values follow the mathematical properties desired, especially orthogonality, or 'perpendicularity', as we would recognize in real-number vector spaces. It transforms the calculations and helps achieve real numbers from complex operations.
Vector Mathematics
In the realm of vector mathematics, vectors are mathematical objects that have both magnitude (length) and direction. When we extend these vectors into the complex domain, each component of the vector is a complex number. This means every calculation, including addition, subtraction, and dot product, can involve complex arithmetic. When working with complex vectors:
  • The dot product, as discussed, involves using complex conjugates to maintain consistent mathematical properties.
  • Magnitudes of vectors and angles between them—conceptually similar to those in real-number vectors—are defined but often require complex algebra techniques.
  • Orthogonality, which refers to vectors being perpendicular, remains crucial, particularly in fields like signal processing and quantum physics.
This mathematical framework allows us to explore deeper relationships and properties within vector spaces.
Orthogonal Vectors in Complex Space
Similar to real-number vector spaces, orthogonality in complex vector spaces implies two vectors are perpendicular. Two vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal if their dot product is zero. The formula for the dot product is applied by including the complex conjugate:
  • \( \mathbf{a} \cdot \mathbf{b} = 0 \).
Given the complex nature of the vector components, this zero condition ensures the vectors occupy positions that are 'perpendicular' in conceptual terms within the complex plane. Orthogonal vectors in complex spaces are fundamental in areas such as quantum mechanics and electromagnetic theory, where the structure of complex spaces provides additional insights into physical phenomena. Understanding their orthogonality helps in decomposing signals into independent components that simplify analysis and processing.

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Most popular questions from this chapter

Think of \(\mathbb{R}^{n}\) as consisting of rows. a. Let \(E\) be an \(n \times n\) matrix, and let \(U=\left\\{\mathbf{x} E \mid \mathbf{x}\right.\) in \(\left.\mathbb{R}^{n}\right\\} .\) Show that the following are equivalent. i. \(E^{2}=E=E^{T}(E\) is a projection matrix). ii. \((\mathbf{x}-\mathbf{x} E) \cdot(\mathbf{y} E)=0\) for all \(\mathbf{x}\) and \(\mathbf{y}\) in \(\mathbb{R}^{n}\) iii. \(\operatorname{proj}_{U} \mathbf{x}=\mathbf{x} E\) for all \(\mathbf{x}\) in \(\mathbb{R}^{n}\). [Hint: For (ii) implies (iii): Write \(\mathbf{x}=\mathbf{x} E+\) \((\mathbf{x}-\mathbf{x} E)\) and use the uniqueness argument preceding the definition of proj \(_{U} \mathbf{x} .\) For (iii) implies (ii): \(\mathbf{x}-\mathbf{x} E\) is in \(U^{\perp}\) for all \(\mathbf{x}\) in \(\left.\mathbb{R}^{n} .\right]\) b. If \(E\) is a projection matrix, show that \(I-E\) is also a projection matrix. c. If \(E F=0=F E\) and \(E\) and \(F\) are projection matrices, show that \(E+F\) is also a projection matrix. d. If \(A\) is \(m \times n\) and \(A A^{T}\) is invertible, show that \(E=A^{T}\left(A A^{T}\right)^{-1} A\) is a projection matrix.

If \(A\) is positive definite, show that \(A=C C^{T}\) where \(C\) has orthogonal columns.

If \(A\) is \(n \times n\) show that \(A A^{T}\) and \(A^{T} A\) are similar. [Hint: Start with an SVD for \(A\).]

Apply the power method to \(A=\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right],\) starting at \(\mathbf{x}_{0}=\left[\begin{array}{l}1 \\\ 1\end{array}\right] .\) Does it converge? Explain.

In each case, find a basis over \(\mathbb{C},\) and determine the dimension of the complex subspace \(U\) of \(\mathbb{C}^{3}\) (see the previous exercise). a. \(U=\\{(w, v+w, v-i w) \mid v, w\) in \(\mathbb{C}\\}\) b. \(U=\\{(i v+w, 0,2 v-w) \mid v, w\) in \(\mathbb{C}\\}\) c. \(U=\\{(u, v, w) \mid i u-3 v+(1-i) w=0 ;\) \(u, v, w\) in \(\mathbb{C}\\}\) d. \(U=\\{(u, v, w) \mid 2 u+(1+i) v-i w=0 ;\) \(u, v, w\) in \(\mathbb{C}\\}\)

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