The Exploration and Exploration Discussion provide students with practical motivation, and insight into common application areas.
Instructions
Within the Module 3 Exploration, after reading through the Exploration tabs, and checking your knowledge in the Check Understanding tab, click on the Discuss! tab. Choose one of the applications, and one of the numbered Discussion Topics for that application to explore with your classmates.
Instructions for Exploration Discussion Posts
- Create a thread that answers the question(s) posed in the Discussion Topic to the best of your ability. Provide as much detail as possible.
- Title your thread with the name of the application and the number of the Discussion Topic (e.g. Crystal Lattices, Discussion Topic #2), and restate the Discussion Topic at the top of your post.
- Take part in the resulting discussions for this Discussion Topic, and at least one other Discussion Topic from your chosen Application, or from one of the other Applications. You must have at least one main post, and at least two other substantive discussion posts to satisfy the requirements for this Exploration Discussion.
Module 03/M3_Exploration.html
Exploration of Vector Spaces and Subspaces
Vector spaces and subspaces
Outcomes
- Recognize the general vector space framework with basic definitions and problems. (D)
- Demonstrate how closely other vector spaces resemble Rn. (D)
- Use vector space terminology to tie together important facts about rectangular matrices. (A, B, D)
- Apply theory to Markov Chains models in biology, business, and engineering. (A, B, C, D, I)
In these Explorations, we will explore a few introductory applications of vector spaces and subspaces. Click on the tabs to explore each application, and read and watch all of the examples. Then, answer the questions on the Check Understanding tab.
The Discuss! tab contains a few discussion questions about the applications in this module, for use in the Exploration Discussion. The Exploration Discussion will provide you with an opportunity to expand further upon the ideas presented here.
Crystal Lattices
Note: If equations don't render correctly when initially viewed, refresh your browser page and wait for it to load completely before visiting any of the tabs.
Some applications have a physically natural coordinate system that may be different from the standard coordinate system with standard unit basis vectors in Rn{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mtext mathvariant="bold">R</mtext><mi>n</mi></msup></math>"}.
In crystallography, a crystal lattice in R3{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mtext mathvariant="bold">R</mtext><mn>3</mn></msup></math>"} is composed of unit cells, which are minimal parallelepipeds, that when translated by each of the lattice vectors along three adjacent edges, make up the crystal lattice, as shown.
Fig. 3.1.1 (generated from): Averill, B., & Eldredge, P. (2007). Chemistry: Principles, patterns, and applications. San Francisco, CA: Pearson Benjamin Cummings.
Figure 3.1.1: A three-dimensional unit cell with its three lattice vectors, and the resulting crystal lattice.
The arrangement of atoms within a unit cell is repeated throughout the crystal lattice. Therefore, knowledge about a crystal lattice can be reduced to knowledge about the unit cell and the arrangement of atoms within it. There are 14 basic types of unit cells comprising all known crystal lattice structures.
Scientists use the lattice vectors that make up the adjacent edges of a unit cell as their working basis of R3{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mtext mathvariant="bold">R</mtext><mn>3</mn></msup></math>"}, forming a natural coordinate system. This gives the atomic arrangement within each unit cell its natural coordinates, whereas using the standard unit basis and resulting unit cubes might be useful for some calculations, but unintuitive for others, since they do not fit the crystal’s natural structure.
By using this natural coordinate system, scientists can combine atom configurations with formulas derived from the relevant type of unit cell to determine such things about a given material as: in what direction it is likely to fracture, where it is easiest to make a clean cut, in what direction it has the greatest tensile strength, or the greatest compressive resistance, and so on. These directions are naturally aligned with the directions of the lattice vectors, so using the lattice vectors as a basis is not only natural, but also a very useful choice.
Example 3.1.1: Finding a Coordinate Vector of an Atom Relative to a Lattice Basis for Graphite
A lattice basis for graphite is given by LB=a3/2-1/20, a3/21/20, c001{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi><mi>B</mi><mo>=</mo><mfenced open="{" close="}"><mrow><mi>a</mi><mfenced open="[" close="]"><mtable><mtr><mtd><msqrt><mstyle displaystyle="false"><mn>3</mn></mstyle></msqrt><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mo>-</mo><mn>1</mn><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mi>a</mi><mfenced open="[" close="]"><mtable><mtr><mtd><msqrt><mn>3</mn></msqrt><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mi>c</mi><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></mrow></mfenced></math>"}, where a=2.46 Ao{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi><mo>=</mo><mn>2</mn><mo>.</mo><mn>46</mn><mo> </mo><mover><mtext>A</mtext><mtext>o</mtext></mover></math>"} (angstroms), and c=6.71 Ao{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>c</mi><mo>=</mo><mn>6</mn><mo>.</mo><mn>71</mn><mo> </mo><mover><mtext>A</mtext><mtext>o</mtext></mover></math>"}.
An atom within a unit cell is located at v=-a/23-a/2c/2{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="bold">v</mtext><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mi>a</mi><mo>/</mo><mn>2</mn><msqrt><mn>3</mn></msqrt></mtd></mtr><mtr><mtd><mo>-</mo><mi>a</mi><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mi>c</mi><mo>/</mo><mn>2</mn></mtd></mtr></mtable></mfenced></math>"}. The following video shows how to find the coordinates of this atom relative to the lattice basis.
Video 3.1.1: Finding the Coordinates of an Atom Relative to a Lattice Basis
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Video by Dr. Lisa Korf: Finding the Coordinates of an Atom Relative to a Lattice Basis [© CCCOnline]
Video 3.1.1 Transcript: Finding the Coordinates of an Atom Relative to a Lattice Basis
It is also important to be able to convert from a lattice basis back to the standard unit basis or to another basis that may be useful for other computations. One such basis is called the reciprocal lattice basis.
Example 3.1.2: Converting from the Lattice Basis for Graphite to the Reciprocal Lattice basis
We begin again with the lattice basis for graphite given by LB=a3/2-1/20, a3/21/20, c001{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>L</mi><mi>B</mi><mo>=</mo><mfenced open="{" close="}"><mrow><mi>a</mi><mfenced open="[" close="]"><mtable><mtr><mtd><msqrt><mn>3</mn></msqrt><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mo>-</mo><mstyle displaystyle="false"><mn>1</mn></mstyle><mstyle displaystyle="false"><mo>/</mo></mstyle><mstyle displaystyle="false"><mn>2</mn></mstyle></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mi>a</mi><mfenced open="[" close="]"><mtable><mtr><mtd><msqrt><mn>3</mn></msqrt><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mi>c</mi><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></mrow></mfenced></math>"}.
This time, we are given a coordinate vector with respect to the lattice basis, vLB=-201{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mfenced open="[" close="]"><mtext mathvariant="bold">v</mtext></mfenced><mrow><mi>L</mi><mi>B</mi></mrow></msub><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math>"}, and we want to know the coordinates of v{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="bold">v</mtext></math>"}
with respect to the reciprocal lattice basis given by RLB=2πa1/3-10, 2πa1/310, 2πc001.
Video 3.1.2: Converting Coordinates Between Two Lattice Bases
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Video by Dr. Lisa Korf: Converting the Coordinates Between Two Lattice Basis [© CCCOnline]
Video 3.1.2 Transcript: Converting Coordinates Between Two Lattice Bases
Crystallography has many areas of application, such as to metallurgy, solid state electronics, and geology. In all of these fields, it is important to be able to perform change of basis operations from one coordinate system to another.
Markov Chains
Consider a system that can be in exactly one of n possible states. Suppose that the system evolves in time such that the probability of being in a particular state at a given time period k depends only on the system's state at the previous time period, k-1{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>k</mi><mo>-</mo><mn>1</mn></math>"}.
For example, suppose you present your pet cat, Cosmo, with a Box Game, in which there are three boxes that she can open. Inside box 1 is a cat-chasing Chihuahua, while box 2 is empty, and box 3 contains an enticing mouse. If Cosmo opens box 1 and finds the Chihuahua, there is only a 2% probability that she will open that box the next time. If she opens box 2, finding it empty, there is a 20% chance she will open that box the next time, and if she opens box 3, there is a 90% chance she will open that box for a repeat game of cat and mouse. We'll assume that Cosmo only remembers what happened the previous time she opened a box, but not any of the times prior to that.
Such a scenario can be modeled with a Markov Chain.
Definition 3.2.1. A probability vector is a vector of states whose entries sum to 1. A stochastic matrix is a square matrix with probability vectors as its columns, i.e. the entries of every column sum to one. A Markov Chain is a sequence of probability vectors x0, x1, …, xk, xk+1, …{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mtext mathvariant="bold">x</mtext><mn>0</mn></msub><mo>,</mo><mo> </mo><msub><mtext mathvariant="bold">x</mtext><mn>1</mn></msub><mo>,</mo><mo> </mo><mo>…</mo><mo>,</mo><mo> </mo><msub><mtext mathvariant="bold">x</mtext><mi>k</mi></msub><mo>,</mo><mo> </mo><msub><mtext mathvariant="bold">x</mtext><mrow><mi>k</mi><mo>+</mo><mn>1</mn></mrow></msub><mo>,</mo><mo> </mo><mo>…</mo></math>"}, together with a stochastic matrix P, such thatx1=Px0, x2=Px1, …, xk+1=Pxk, ….The probability vectors represent a sequence of states, and the stochastic matrix P has entries Pij that give the transition probabilities of going from state j to state i.
Example 3.2.1: The Stochastic Matrix and Probability Vectors for the Cat and Box Game
If Cosmo the cat picks box 1 on her first try, then the probability that she will pick box 1 on her second try is only 2%, so the stochastic matrix P for this example would have entry P11=0.02. Assuming that the remaining probability of 0.98 is split evenly between the remaining boxes in that column, and similarly for the remaining columns, we can write down the stochastic matrix for Cosmo: From Box: P= 1 2 3 0.020.40.050.490.20.050.490.40.9To Box:123If Cosmo has a 1/3 probability of opening each of the three boxes at time period 0, her initial state vector would be x0=1/31/31/3{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mtext mathvariant="bold">x</mtext><mn>0</mn></msub><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn><mo>/</mo><mn>3</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>/</mo><mn>3</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>/</mo><mn>3</mn></mtd></mtr></mtable></mfenced></math>"}. The following video computes the first 6 state vectors that result after Cosmo chooses which box to open at 6 consecutive time periods.
Video 3.2.1: Computing the First 6 State Vectors of a Markov Chain
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Video by Dr. Lisa Korf: Computing the First 6 State Vectors of a Markov Chain [© CCCOnline]
Video 3.2.1 Transcript: Computing the First 6 State Vectors of a Markov Chain
From the video, it appears that as Cosmo continues to choose boxes, her state vector starts to hone in on a single vector. This is not a coincidence, due to the following theorem.
Definition 3.2.2. A stochastic matrix P is regular if there is any power of P whose entries are all strictly positive.
Theorem 3.2.1. If P is a regular stochastic matrix, then there is a unique probability vector, q, such that Pq = q. The vector q is called the steady state vector for P. In addition, no matter what the initial probability vector x0 is, the sequence of probability vectors {xk} will converge to q as k tends to infinity.
Example 3.2.2: The Steady State Vector for the Cat and Box Game
To find a steady state vector q, we know from the theorem above that it must be the unique probability vector that solves the equation Px=x. So let's use linear algebra to find Cosmo's steady state vector!
Video 3.2.2: Finding the Steady State Vector
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Video by Dr. Lisa Korf: Finding the Steady State Vector [© CCCOnline]
Video 3.2.2 Transcript: Finding the Steady State Vector
We can now see that in the long run, Cosmo will pick box 1 with probability 40/481, or about 8.316% of the time, box 2 with probability 49/481, or about 10.187% of the time, and box 3 with probability 392/481, or about 81.497% of the time. Based on these results, do you think that in reality, Cosmo's memory would really only go back one time period, or would she be more likely to avoid box 1, perhaps indicating a longer-term memory?
Image Attributions: Dog Silhouette: https://pixabay.com/en/dog-puppy-animal-black-silhouette-159481/ , Cat Silhouette: http://www.clipartbest.com/clipart-ndT88AKie , Mouse Silhouette: (mirrored image) http://www.rgbstock.com/bigphoto/mSdQGMK/silhouette+gerbil
check your understanding
Crystal Lattices:
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Suppose that a crystal lattice has a lattice basis given by LB = { u, v, w }, and that an atom is located at x = u + 3v + w. What are the coordinates of the atom with respect to the lattice basis LB?
See Answer
Answer: xLB=131{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mfenced open="[" close="]"><mtext mathvariant="bold">x</mtext></mfenced><mrow><mi>L</mi><mi>B</mi></mrow></msub><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>3</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math>"}
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A crystal lattice has a particular atom type appearing at coordinates (1, 2, –3), and repeating in three different directions at coordinates (–1, 2, 4), (2, 3, 5), and (4, –2, 1). What is a possible lattice basis for this crystal lattice?
See Answer
Answer: -1-12-24-(-3), 2-13-25-(-3), 4-1-2-21-(-3)=-207, 118, 3-44{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="{" close="}"><mrow><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mn>1</mn><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>2</mn><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>4</mn><mo>-</mo><mo>(</mo><mo>-</mo><mn>3</mn><mo>)</mo></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>2</mn><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>3</mn><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>5</mn><mo>-</mo><mo>(</mo><mo>-</mo><mn>3</mn><mo>)</mo></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>4</mn><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mo>-</mo><mn>2</mn><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>-</mo><mo>(</mo><mo>-</mo><mn>3</mn><mo>)</mo></mtd></mtr></mtable></mfenced></mrow></mfenced><mo>=</mo><mfenced open="{" close="}"><mrow><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>7</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>8</mn></mtd></mtr></mtable></mfenced><mo>,</mo><mo> </mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>3</mn></mtd></mtr><mtr><mtd><mo>-</mo><mn>4</mn></mtd></mtr><mtr><mtd><mn>4</mn></mtd></mtr></mtable></mfenced></mrow></mfenced></math>"}
Markov Chains 
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Suppose that Cosmo's initial probability vector had for some reason been 001{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math>"}, with all other probabilities the same. How would this affect Cosmo's steady state vector?
See Answer
Answer: It wouldn't, since when a steady state vector exists, which it does in this case, it is independent of the initial probability vector state.
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For a Markov chain with stochastic matrix equal to the 2 by 2 matrix, P=0110{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mn>0</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>0</mn></mtd></mtr></mtable></mfenced></math>"}, does a steady state vector exist? If so, what is it? If not, why not?
See Answer
Answer: Although this matrix does not satisfy the assumptions of the theorem, such a matrix may still have a steady state if it has a unique probability vector solution to the equation P-Ix=0{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><mi>P</mi><mo>-</mo><mi>I</mi></mrow></mfenced><mtext mathvariant="bold">x</mtext><mo>=</mo><mn mathvariant="bold">0</mn></math>"}. In this case, P-I=-111-1{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>P</mi><mo>-</mo><mi>I</mi><mo>=</mo><mfenced open="[" close="]"><mtable><mtr><mtd><mo>-</mo><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mo>-</mo><mn>1</mn></mtd></mtr></mtable></mfenced></math>"}, which has reduced row-echelon form, 1-100{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn></mtd><mtd><mo>-</mo><mn>1</mn></mtd></mtr><mtr><mtd><mn>0</mn></mtd><mtd><mn>0</mn></mtd></mtr></mtable></mfenced></math>"}. The solution to the homogeneous system is x=x211{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mtext mathvariant="bold">x</mtext><mo>=</mo><msub><mi>x</mi><mn>2</mn></msub><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd></mtr></mtable></mfenced></math>"}. Letting x2=1/2{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mn>2</mn></msub><mo>=</mo><mn>1</mn><mo>/</mo><mn>2</mn></math>"} yields the unique probability vector solution, 1/21/2{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="[" close="]"><mtable><mtr><mtd><mn>1</mn><mo>/</mo><mn>2</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>/</mo><mn>2</mn></mtd></mtr></mtable></mfenced></math>"}, which is also the steady state vector.
Discuss!
Choose one of the applications, and one of the numbered Discussion Topics for that application to explore with your classmates in the Exploration Discussion in this Module.
Crystal Lattices – Discussion Topics
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Using the lattice basis LB and the reciprocal lattice basis RLB given for graphite in Examples 3.1.1 and 3.1.2, you wish to determine the location of a point of interest within