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Find the lengths and the inner product of $$ x=\left[\begin{array}{c} 2-4 i \\ 4 i \end{array}\right] \quad \text { and } \quad y=\left[\begin{array}{c} 2+4 i \\ 4 i \end{array}\right] $$

Short Answer

Expert verified
Lengths are 6 each, inner product is 4 - 16i.

Step by step solution

01

Find the Length of x

To find the length of the vector \( x \), we use the formula for the norm: \[\| x \| = \sqrt{x_1 \overline{x_1} + x_2 \overline{x_2}}\]Plugging in the components of \( x \): \[x = \begin{bmatrix} 2 - 4i \ 4i \end{bmatrix}\]we have:\[x_1 = 2 - 4i \quad \text{and} \quad x_2 = 4i\]Calculating the conjugates:\[\overline{x_1} = 2 + 4i \quad \text{and} \quad \overline{x_2} = -4i\]Thus, the norm is:\[\| x \| = \sqrt{(2 - 4i)(2 + 4i) + (4i)(-4i)}\]\[= \sqrt{(4 + 16) + 16}\]\[= \sqrt{36} = 6.\]
02

Find the Length of y

Similarly, we find the length of vector \( y \):\[y = \begin{bmatrix} 2 + 4i \ 4i \end{bmatrix}\]Here:\[y_1 = 2 + 4i \quad \text{and} \quad y_2 = 4i\]Calculating the conjugates:\[\overline{y_1} = 2 - 4i \quad \text{and} \quad \overline{y_2} = -4i\]Thus, the norm is:\[\| y \| = \sqrt{(2 + 4i)(2 - 4i) + (4i)(-4i)}\]\[= \sqrt{(4 + 16) + 16}\]\[= \sqrt{36} = 6.\]
03

Calculate the Inner Product of x and y

The inner product of two vectors \( x \) and \( y \) is given by:\[x \cdot y = x_1 \overline{y_1} + x_2 \overline{y_2}\]Substitute the components:\[x \cdot y = (2 - 4i)(2 - 4i) + (4i)(-4i)\]Calculate each term:The first term:\[(2 - 4i)(2 - 4i) = 4 - 8i - 8i - 16 = -12 - 16i\]The second term:\[(4i)(-4i) = -16(-1) = 16\]Combine:\[x \cdot y = (-12 - 16i) + 16 = 4 - 16i\]Thus, the inner product is \(4 - 16i\).

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Key Concepts

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

Complex Numbers
Complex numbers play a fundamental role in various fields of mathematics and engineering. They are composed of a real part and an imaginary part. The general form of a complex number is expressed as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
The imaginary part includes the imaginary unit \(i\), which is equal to the square root of -1. This means \(i^2 = -1\). In operations involving complex numbers, special rules apply like handling the \(i\) component.
When working with complex numbers in vector-related problems, each element of a vector could be a complex number. In our exercise, for instance, vectors \(x\) and \(y\) have components like \(2 - 4i\) and \(2 + 4i\) respectively.
These numbers depict how imaginary numbers blend with real numbers in higher mathematics to enable the handling of rotations, oscillations, and even abstract mathematical concepts. When performing calculations like norms or inner products, the properties of complex numbers, particularly the handling of \(i\), become crucial.
  • Real Part: The non-imaginary component of the complex number.
  • Imaginary Part: The component involving \(i\).
  • Standard Form: \(a + bi\).
Vector Norm
A vector norm is a measure of a vector's length or magnitude. For vectors composed of complex numbers, the calculation adapts to account for the complex nature of the elements. The norm of a vector \(x\) with complex components is computed using:
\[||x|| = \sqrt{x_1 \overline{x_1} + x_2 \overline{x_2}}\]
This formula utilizes the components of the vector and their conjugates, ensuring the measure is appropriately adjusted for any complex numbers. In essence, to find a norm, we pair each component with its conjugate, multiply, and sum those values before taking the square root.
In our solution, the norm for both vectors \(x\) and \(y\) was found to be \(|x| = 6\) and \(|y| = 6\). This reveals that both vectors have the same magnitude in this complex dimensional space.
  • Normalization accounts for real and imaginary parts.
  • Ensures accurate measurements for complex-related data.
  • Combined conjugate components to handle complex arithmetic.
Conjugate
The conjugate of a complex number involves flipping the sign of its imaginary part. If a complex number is expressed as \(a + bi\), its conjugate is \(a - bi\).
This operation is pivotal in complex mathematics because it facilitates various calculations, including those for norms and inner products. Specifically, using conjugates ensures that operations correctly handle the imaginary component, aiding in producing real number outputs where needed.
In the solution process for the exercise, conjugates were used to handle the vector components during norm calculations and the inner product. For example, for \(x_1 = 2 - 4i\), the conjugate is \(2 + 4i\). Similarly, for \(x_2 = 4i\), the conjugate is \(-4i\).
The art of using conjugates effectively aligns the computations to result in a coherent and accurate mathematical representation, vital in vector spaces involving complex numbers.
  • Flips the sign of imaginary parts
  • Essential for inner and dot products
  • Integral in forming real outputs from complex data

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Most popular questions from this chapter

Find the largest \(a, b, c\) for which these matrices are stable or neutrally stable: $$ \left[\begin{array}{rr} a & -.8 \\ 8 & .2 \end{array}\right], \quad\left[\begin{array}{ll} b & .8 \\ 0 & .2 \end{array}\right], \quad\left[\begin{array}{rr} c & .8 \\ .2 & c \end{array}\right] $$

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Multiplying term by term, check that \((I-A)\left(I+A+A^{2}+\cdots\right)=I\). This series represents \((I-A)^{-1}\). It is nonnegative when \(A\) is nonnegative, provided it has a finite sum; the condition for that is \(\lambda_{\max }<1 .\) Add up the infinite series, and confirm that it equals \((I-A)^{-1}\), for the consumption matrix $$ A=\left[\begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \quad \text { which has } \lambda_{\max }=0 $$

If \(v_{1}, \ldots, v_{n}\) is an orthonormal basis for \(C^{n}\), the matrix with those columns is a matrix. Show that any vector \(z\) equals \(\left(v_{1}^{\mathrm{H}} z\right) v_{1}+\cdots+\left(v_{n}^{\mathrm{H}} z\right) v_{n}\),

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