Chapter 6: Problem 29
Sketch the unit circle in \(R^{2}\) using the given inner product. $$\langle\mathbf{u}, \mathbf{v}\rangle=\frac{1}{4} u_{1} v_{1}+\frac{1}{16} u_{2} v_{2}$$
Short Answer
Expert verified
The unit circle is an ellipse: \(\frac{x_1^2}{4} + \frac{x_2^2}{16} = 1\).
Step by step solution
01
Understand the Unit Circle
The equation of a unit circle in standard form with an inner product is \( \langle \mathbf{x}, \mathbf{x} \rangle = 1 \). This condition ensures the "length" or magnitude of the vector \( \mathbf{x} \) in the space defined by \( \langle \cdot, \cdot \rangle \) is 1.
02
Apply the Inner Product
Using the given inner product \( \langle \mathbf{x}, \mathbf{x} \rangle = \frac{1}{4}x_1^2 + \frac{1}{16}x_2^2 \), the unit circle equation becomes \( \frac{1}{4}x_1^2 + \frac{1}{16}x_2^2 = 1 \).
03
Simplify the Equation
To express the equation of the unit circle in a more familiar form, multiply each term by 16 to clear the fractions: \( 4x_1^2 + x_2^2 = 16 \).
04
Recognize as an Ellipse
The equation \( \frac{x_1^2}{4} + \frac{x_2^2}{16} = 1 \) is recognizable as an ellipse. It is not a traditional circle but an ellipse stretched along the \(x_2\) direction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inner Product
An inner product is a mathematical concept that allows us to determine an angle or a "length" in a vector space. It generalizes the dot product and introduces a way to measure similarity between two vectors in more complex spaces. In the problem you've encountered, the inner product \[\langle \mathbf{u}, \mathbf{v} \rangle = \frac{1}{4} u_{1} v_{1} + \frac{1}{16} u_{2} v_{2}\]gives us a specific way to compute this, with coefficients \( \frac{1}{4} \) and \( \frac{1}{16} \) that weight the contributions of different vector dimensions.
Key points about inner products:
Key points about inner products:
- They provide a measure of the vector's "length" and can influence its geometry.
- They are often symmetric, meaning \( \langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle \).
- The inner product of a vector with itself relates to its magnitude or "norm" based on the specific properties of the space.
Ellipse
In geometry, an ellipse is a curve on a plane that surrounds two focal points. In our context, with the given inner product, the so-called unit circle is not a circle but rather an ellipse. This arises from how different dimensions (\(x_1\) and \(x_2\) in our case) contribute unevenly to the vector's magnitude.
The equation for our ellipse is \[\frac{x_1^2}{4} + \frac{x_2^2}{16} = 1\]This structure shows that the ellipse is wider along the \(x_2\)-axis.
Ellipses have significant properties:
The equation for our ellipse is \[\frac{x_1^2}{4} + \frac{x_2^2}{16} = 1\]This structure shows that the ellipse is wider along the \(x_2\)-axis.
Ellipses have significant properties:
- They are symmetrical about two axes through their center.
- The major axis is longer than the minor axis, stretching according to the weight each dimension carries in the inner product.
- In our exercise, the axes’ lengths are determined by the inverse of the weights from the inner product.
Magnitude of a Vector
The magnitude of a vector is akin to the length of a vector from its initial point to its terminal point. In standard Euclidean space, you might calculate this as \(\sqrt{x_1^2 + x_2^2}\). However, in our specific problem, the inner product defined a different way to evaluate magnitude.
To ensure the vector \(\mathbf{x}\) has a magnitude of 1 — meaning it lies on what we call the "unit circle" — we used our inner product \[\langle \mathbf{x}, \mathbf{x} \rangle = \frac{1}{4} x_1^2 + \frac{1}{16} x_2^2 = 1\]The coefficients \( \frac{1}{4} \) and \( \frac{1}{16} \) alter how the magnitude is computed, effectively scaling the contribution of each component.
Key observations about vector magnitude:
To ensure the vector \(\mathbf{x}\) has a magnitude of 1 — meaning it lies on what we call the "unit circle" — we used our inner product \[\langle \mathbf{x}, \mathbf{x} \rangle = \frac{1}{4} x_1^2 + \frac{1}{16} x_2^2 = 1\]The coefficients \( \frac{1}{4} \) and \( \frac{1}{16} \) alter how the magnitude is computed, effectively scaling the contribution of each component.
Key observations about vector magnitude:
- Magnitude is always a non-negative scalar showing the vector's "size".
- Within different vector spaces, as defined by inner products, the magnitude can behave differently.
- This scalar provides insight into how much different dimensions contribute, given the inner product's influence on geometric configurations.