8000 GitHub - hhdsml/1806-algebra: 18.06 course at MIT
[go: up one dir, main page]
More Web Proxy on the site http://driver.im/
Skip to content

hhdsml/1806-algebra

BranchesTags
 
 

Folders and files

NameName
Last commit message
Last commit date

Latest commit

 

History

2,350 Commits
exams
exams
 
 
notes
notes
 
 
psets
psets
 
 
.gitignore
.gitignore
 
 
README.md
README.md
 
 

Repository files navigation

MIT Course 18.06, Fall 2022

This is a repository for the course 18.06: Linear Algebra at MIT in Fall 2022. See other branches of this repository for previous semesters.

Instructor: Prof. Steven G. Johnson. Course administrator: Sergei Korotkikh.

Lectures: MWF11 in 26-100. Handwritten notes are posted online, along with video recordings (on Canvas Panopto Video) and other materials (slides, further reading) in the lecture summaries below.

Exams: 11am in 26-100, on 10/7, 11/14, & 12/9. Final exam: 12/22 9am-noon in Dupont Gym.

Recitations:

Undergraduate Assistants: Raza Abbas, Alvin Chen, Christina Mirro, and Daniel Villagran. Email them at 1806fall22_ua ατ mit.edu for 1-on-1 technical help with Julia or other questions that don't work well over Piazza etc.

Resources: Piazza discussion forum, math learning center, TSR^2 study/resource room, pset partners.

This document is a brief summary of what was covered in each 18.06 lecture, along with links and suggestions for further reading. It is not a good substitute for attending lecture, but may provide a useful study guide. (You can also look at the analogous summaries from Spring 2022.)

Lecture 1 (Sep 7)

Slides giving the syllabus and the "big picture" of what 18.06 is about. Introduction to thinking about matrices as linear operations, not just as "bags of numbers".

Further reading: Strang, chapter 1, and section 8.1 on linear transformations. 3blue1brown has a nice video on matrix multiplication as composition of linear transformations. If you've forgotten the basics of how to multiply matrices by vectors or matrices by matrices, google for some tutorial material online (e.g. Khan academy) and do a quick brush-up.

Lecture 2 (Sep 9)

  • handwritten notes: see link above (at beginning)
  • video

Gaussian elimination for Ax=b: I started with the grade-school/high-school viewpoint of writing out three equations in three unknowns, adding/subtracting multiples of equations until we were left with one equation in one unknown. Then, I wrote the same equations in matrix form, and renamed this process "Gaussian elimination": we add/subtract multiples of matrix rows to introduce zeros below the diagonal, i.e. to make the matrix upper triangular U. We then do the same row operations to the right hand side b to get a new vector c. Finally, we solve Ux=c for x by working from bottom (1 equation in 1 variable) to top, a process called "backsubstitution".

To do the same operations to both A and b, a useful trick for hand calculations is to augment the matrix with a new column representing the right-hand side, forming [A b] before starting Gaussian elimination.

What comes next? The problem with expressing Gaussian elimination this way, as operations on individual numbers in the matrix, is that it is impossible to follow the process in detail for anything except a very tiny matrix. We need a higher-level "algebraic" way to express the process, both to help us understand it and also to help us to use it (e.g. to perform additional algebraic transformations afterwards). To do this, we want to express the process, not as operations on individual numbers, but as matrix operations.

Rewrote Gaussian elimination in matrix form: we multiply a matrix A on the left by a sequence of lower-triangular "elimination matrices" Eₙ to arrive at an upper-triangular matrix U = EA. To solve Ax=b, we can think of the earlier process as multiplying both sides on the left by E, the linear operator representing the composition (product) of all of the elimination steps, yielding Ux=EAx on the left and c=Eb on the right.

We're not done: it turns out to be even more fruitful to reverse the process, and write A = LU: L represents the operations required to turn the matrix U back into A, and turns out toe be a lower-triangular matrix whose entries are just a record of the elimination steps. This LU factorization is extremely useful and important because it allows us to replace a complicated matrix A with two much simpler (triangular) ones. For example, solving Ax=b turns into LUx=b, and we can do this just by two "triangular" solves. More on this next time.

Further reading: Textbook sections 2.1, 2.2, 2.3. Strang lecture 2 video. And there is a Gaussian-elimination Julia notebook that covers the same steps in Julia form. See also "The key reason why A = LU" in section 2.6 of the textbook.

Lecture 3 (Feb 4): recorded

(Prof. Johnson is sick and so we will use the recorded lecture from spring.)

Showed that Gaussian elimination can be viewed as LU factorization:

  • Gaussian elimination A ⟿ U=EA (without row swaps) can be thought of as A=LU: factorizing A into a product of two simpler (triangular) matrices (L=lower, U=upper). U is the matrix that you normally get when you do elimination by hand, and L (the inverse of the elimination steps L=E⁻¹, a lower-triangular matrix with 1's on the diagonal) is essentially a "record" of the elimination steps.

L is the matrix that "reverses" Gaussian elimination: it tells you how to get A back from L. Despite this, I showed in lecture that L is actually easier to get than E: all you do is make a diagonal matrix of 1's, and then fill in the multipliers from the elimination steps (flipping subtraction to addition) below the diagonal. So, L just requires bookkeeping, and no computation.

Computing U is hard (elimination is a lot of work, even for a computer), but once you have U and L then many things that you might want to do with A become easy.

  • For example, suppose you want to solve Ax=b, given A=LU. Write LUx=b=L(Ux), and let y=Ux. Then Ly=b, and we can solve for y by forward-substitution. Given y, we can then solve Ux=y by back-substitution. Both of these steps are easy because the systems are triangular.
    • Moreover, solving Ly=b turns out to exactly correspond to applying the elimination steps from A ⟿ U to b. (The 1's on L's diagonal mean that there are no divisions required, either.)
  • This means that we can re-use L and U to solve Ax=b for many right-hand sides. In contrast, if you "augment" A with b and then do elimination (A|b)⟶(U|y), you get the same new right-hand side y but you haven't kept a record of the elimination steps, so if you have a new right-hand side you might naively repeat the whole elimination process (hard!) rather than solving Ly=b (easy!).
  • More generally, whenever you have A as a product of "simpler" matrices (e.g. triangular, diagonal, …), you can solve Ax=b by a sequence of "simpler" solves.

Introduction to the concept of a matrix inverse more generally as the matrix that reverses the action of a linear operator. Key ideas from the notebook:

  • A⁻¹ is the matrix that does the "reverse" of A: A⁻¹(Ax)=x for any x. It also follows that A is the reverse of A⁻¹: A(A⁻¹x)=x for any x, i.e. (A⁻¹)⁻¹=A.
    • That is, if Ax=b, then A⁻¹b=x (for any x). (Equivalently, it gives the solution to Ax=b.)
    • It only exists for square, nonsingular matrices A. (i.e. an m×m matrix A must give m nonzero pivots when you do elimination.)
  • Equivalently, A⁻¹ is the matrix for which A⁻¹A = A⁻¹A = I (the m×m identity matrix).
    • I is an identity matrix, the matrix that gives Ix=x for any x or IA=A and AI=A for any A. There are m×m identity matrices for all m, and when we write "I" we usually infer from context how big an I we mean.

In the next lecture (which will start with the end of this notebook), we will look at calculating inverses more generally (although it turns out that this is something that you should almost never do explicitly, even on a computer!).

Further reading: Textbook sections 2.5, 2.6. Strang lecture 4 video and lecture 3 video. See also "The key reason why A = LU" in section 2.6 of the textbook.

Optional Julia Tutorial: recorded

(No live tutorial since Prof. Johnson is sick.)

A basic overview of the Julia programming environment for numerical computations that we will use in 18.06 for simple computational exploration. This (Zoom-based) tutorial will cover what Julia is and the basics of interaction, scalar/vector/matrix arithmetic, and plotting — we'll be using it as just a "fancy calculator" and no "real programming" will be required.

If possible, try to install Julia on your laptop beforehand using the instructions at the above link. Failing that, you can run Julia in the cloud (see instructions above).

Lecture 4 (Sep 14)

Reviewed matrix inverses and key properties thereof.

Went through how to explicitly compute A⁻¹ by solving AA⁻¹ = I. Essentially, this is just solving Ax=b multiple times, where b is each column of I. A common way to organize this for hand calculation (ugh) is the Gauss–Jordan algorithm (on a 3×3 example that can also be found in the textbook): If we do row operations on A to get I, then the same row operations on I give A⁻¹! To carry this out by hand, we augment (A|I), do ordinary Gaussian elimination to get (U|C), and then do elimination "upwards" to get (I|A⁻¹).

Matrix inverses are mainly a conceptual tool that we use to move matrices around symbolically in equations. Once you are through with your algebraic manipulations, you might end up with an expression like A⁻¹b — but when it comes time to actually compute the answer, you should read "A⁻¹b" as "solve Ax=b for x by the best available method".

Further reading: Textbook sections 2.5, 2.6. Strang video lecture 3.

Lecture 5 (Sep 16)

Brief review of previous topics in LU factorization with some more examples in the notebook:

  • How the L matrix entries are just the multipliers from Gaussian elimination. No extra work is required!
  • How in practice, one rarely "augments" the matrix with the right-hand side. Instead, you compute A=LU, substitute this into Ax=b=LUx, let c=Ux, solve Lc=b, then solve Ux=c. In particular, solving Lc=b is exactly the same as performing the Gaussian-elimination steps on c. (The "augmented" method is a little easier for human bookkeeping, but has essentially no advantage for the computer.)

Some new information about LU to complete the story:

  • Given A=LU, you can efficiently solve multiple right-hand sides, or equivalently the matrix equation AX=B.
  • How row swaps lead to the factorization PA=LU: in practice, the computer almost always does row swaps, and always gives you a permutation matrix P (or its equivalent).

We apply PA=LU to Ax=b in much the same way as for LU; the only difference is that we have to first apply the permutation P to b.

Permutation matrices P are a great example of a linear operator that is often easier to understand (and more efficient) if you don't write it as a matrix, but instead write it as a "vector" p of the permuted indices 1…n in the new order. Then Px is just x[p] in Julia (and very similarly in Matlab and Numpy): just make a new vector by extracting the components p₁,p₂,… of x.

Further reading: Textbook sections 2.7 (on permutations; we will talk about transposes soon), and 11.1. Strang video lecture 4 and video lecture 5. For 18.06, I don't expect you to know the details of how the permutation P in PA=LU is constructed even though you don't know the permutation in advance … you only need to know how to use PA=LU if it (or something similar) is given to you … but if you are interested this "partial pivoting" algorithm is described in lecture 21 of Numerical Linear Algebra by Trefethen and Bau, or in many other textbooks on numerical linear algebra.

Lecture 6 (Sep 19)

Complexity of matrix operations: why matrix × vector or backsubstitution scale like n² for n×n matrices, while matrix × matrix or Gaussian elimination (LU factorization) scale like n³. Matrices much bigger than a few thousand square quickly become impractical, and really large problems are only tractable because they have special structure like sparsity.

The next few weeks will be devoted to problems arising from singular and non-square matrices. A "singular" square matrix is one for which we "run out of pivots" when doing elimination (we hit zeros on the diagonal we can't remove with row swaps). Non-square matrices are either "tall" (arising in overdetermined systems with more equations than unknowns, leading to fitting problems and approximate solutions) or "wide" (arising in underdetermined systems with more unknowns than equations, leading to freedom in the choice of solution and regularization techniques to impose "priors" on this choice). A key technique that will help us understand these equations is to break vectors into simpler/smaller vectors. To do this we must first broaden our concept of a "vector".

Introduced vector spaces (informally, a set V of anything you can add x±y and multiply by scalars αx) and subspaces (a subset of V such that vector operations stay in the subspace). Examples of vector spaces include real n-component vectors (ℝⁿ, or ℂⁿ for complex numbers), real m×n matrices, functions f(x) (ℝ↦ℝ, real numbers to real numbers). Examples of subspaces includes planes or lines through the origin in ℝ³, or the origin itself. The goal of this is to break vector spaces into smaller pieces that we can still do linear algebra on (hence the need for a subspace, not just any arbitrary subset). Subspaces are especially important to help us understand the solutions (if any) of Ax=b for non-square matrices A.

Further reading: Textbook sections 2.6 ("The cost of elimination") and 11.1. Section 3.1 and 3.2 on vector spaces and subspaces.

Lecture 7 (Sep 21)

  • video: Panopto Video link on Canvas
  • Handwritten notes

Reviewed vector spaces (informally, a set V of anything you can add x±y and multiply by scalars αx) and introduced subspaces (a subset of V such that vector operations stay in the subspace). Examples of vector spaces include real n-component vectors (ℝⁿ, or ℂⁿ for complex numbers), real m×n matrices, functions f(x) (ℝ↦ℝ, real numbers to real numbers). Examples of subspaces includes planes or lines through the origin in ℝ³, or the origin itself. The goal of this is to break vector spaces into smaller pieces that we can still do linear algebra on (hence the need for a subspace, not just any arbitrary subset). Subspaces are especially important to help us understand the solutions (if any) of Ax=b for non-square matrices A.

For an m×n matrix A, introduced two important subspaces.

  • First, the column space C(A): the set {Ax for all x ∈ ℝⁿ}. This is the set of right-hand sides b such that Ax=b is solvable, and is a subspace of the "output space" ℝᵐ (not ℝⁿ unless m=n!). Equivalently, C(A) is all linear combinations of the columns of A, which we call the span of the columns.

  • Second, the null space N(A) (also called the "kernel"): the set {x such that Ax=0} ⊆ ℝⁿ (i.e., a subspace of the "input space" ℝⁿ). Given any solution x to Ax=b, then x+z is also a solution if z ∈ N(A) (i.e. Az=0 ⟹ A(x+z)=Ax+Az=Ax=b).

These are very important subspaces because they tell us a lot about the matrix A and solutions to Ax=b. As a trivial example, if A is an n×n invertible matrix, C(A)=ℝⁿ and N(A)={0}. Conversely, if A is an m×n matrix of zeros, then C(A)={0} and N(A)=ℝⁿ.

Defined a basis for a vector space as a minimal set of vectors (we will later say that they have to be linearly independent) whose span (all linear combinations) produces everything in the space. The number of vectors in any basis is the dimension of the vector space.

Showed that the nullspace is preserved by elimination (row) operations, but that the column space is not. So, to find N(A), we can instead do elimination and find N(U)=N(A) for the upper-triangular form U. We now want to find all possible solutions to Ax=0.

Further reading: Textbook, sections 3.1—3.3; Strang video lecture 6 and lecture 7. Note that Strang's lectures and book emphasize the "reduced row echelon" ("rref") form, which is essentially a bookkeeping trick (similar to Gauss–Jordan for inverses) to do the back-solves for the special solutions all at once. I will not emphasize rref form this semester, but you can use it if you want. (In practical applications, rref form is virtually never used, and for that matter one doesn't actually use elimination at all to find null spaces; instead, one uses something called the SVD that we will cover later.)

Lecture 8 (Sep 26)

Went through two examples of finding the special solutions as a basis for the nullspace. They key point is always this: given the free variables, we can easily solve for the corresponding pivot variables.

We can now find the complete solution (i.e., all solutions) to a non-square linear system Ax=b. Elimination turns this into Ux=c. We look for solutions in the form xₚ + xₙ: a particular solution xₚ plus any vector xₙ in N(A) (specified explicitly from the basis). The particular solution xₚ can be any solution to Ax=b, but the simplest one to find is usually to set the free variables to zero. That is, we write the solution xₚ=(p; 0) where p is the unknown values in the pivot rows, setting the other (free) rows to zero, then plug into Uxₚ=c to get p. Since this extracts the part of U that is upper triangular, we can easily solve it. Then we add in anything in N(A).

I used a 3×4 example matrix A (similar in spirit to one in the textbook, section 3.3) that was rank deficient: after elimination, we only had two pivots (rank r=2) in the first two rows, and a whole row of zeros. Furthermore, its pivot columns were the first and third columns, with the second and fourth columns being free — this is possible (albeit unusual)! Showed that we could still get the special solutions by solving for the pivot variables in terms of the free variables = (1,0,…) etcetera, and we still got dimension n-r (= 2) for the null space. When solving Ax=b by elimination into Ux=c, however, we only got a solution x if c was zero in the same rows as U. If the zero third row of U was matched by a zero third row of c, then we got a particular solution as before by setting the free variables to zero. If the zero third row of U was not matched by a zero third row of c, then there is no solution: b was not in the column space C(A). This gives an easy way to check whether the right-hand-side is in the column space. Went through an example right-hand-sides with no solutions, and in next lecture I'll cover one with infinitely many solutions.

Further reading: Textbook sections 3.3–3.4, lecture 8. As noted in lecture 7, I'm not emphasizing the trick of "rref" form (used extensively by Strang's book and lectures) which makes hand calculation slightly easier, but might push students towards memorizing formulas (which are useless in the long run).

Lecture 9 (Sep 28)

  • handwritten notes and lecture video (see links above).

Finished the discussion of "complete solutions" of Ax=b from last lecture, by choosing a right-hand side b ∈ C(A), finding the particular solution (by setting the free variables to zero and solving for the pivot variables), and then adding in the null-space vectors to describe all possible solutions.

Linear independence and C(A)

Bases, dimension, and independence. Earlier, I defined a basis as a minimal set of vectors whose span gives an entire vector space, and the dimension of the space as the size of the basis. Now, we want to think more carefully about the term "minimal". If we have too many vectors in our basis, the problem is that some of the vectors might be redundant (you can get them from the other basis vectors). We now rephrase this as saying that such vectors are linearly dependent: some linear combination (with nonzero multipliers) of them gives the zero vector, and we want every basis to be linearly independent. The dimension of a subspace is still the number of basis vectors.

What does it mean to be linearly independent? Given a set of n vectors {x₁, ⋯, xₙ}, if we matrix a matrix X whose columns are x₁⋯xₙ, then C(X) is precisely the span of x₁⋯xₙ. To check whether the x₁⋯xₙ form a basis for C(X), we need to check whether they are linearly independent. There are three equivalent ways to think about this:

  1. We want to make sure that none of x₁⋯xₙ are "redundant": make sure that no xⱼ can be made from a linear combination of the other xᵢ's.

  2. Equivalently, we don't want any linear combination of x₁⋯xₙ to give zero unless all the coefficients are zero.

  3. Equivalently, we want N(X) = {0}.

In this way, we reduced the concept of independence to thinking about the null space. Since the nullspace is preserved by elimination, it follows that columns of A are dependent/independent if and only if the corresponding columns of R are dependent/independent. By looking at R, we can see by inspection that the pivot columns form a maximal set of independent vectors, and hence are a basis for C(R). Hence, the pivot columns of A (i.e. the columns of A corresponding to the columns of R or U where the pivots appear) are a basis for C(A).

It follows that the dimension of C(A) is exactly rank(A).

Four important cases for Ax=b

Went through four important cases for an m×n matrix A of rank r. (Note that we must have r ≤ m and n: you can't have more pivots than there are rows or columns.)

  1. If r=n, then A has full column rank. We must have m ≥ n (it is a "tall" matrix), and N(A)={0} (there are no free columns). Hence, any solution to Ax=b (if it exists at all) must be unique. (If m > n, the problem is "overdetermined": more equations than unknowns.)

  2. If r=m, then A has full row rank. We must have n ≥ m (it is a "wide" matrix), and C(A)=ℝᵐ. Ax=b is always solvable (but the solution will not be unique unless m=n). (If m < n the problem is "underdetermined": more unknowns than equations.)

  3. If r=m=n, then A is a square invertible matrix. Ax=b is always solvable and the solution x=A⁻¹b is unique.

  4. If r < m and r < n, then A is rank deficient. Solutions to Ax=b may not exist and will not be unique if they do exist.

Cases (1)-(3) are called full rank: the rank is as big as possible given the shape of A. In practice, most matrices that one encounters are full rank (this is essentially always true for random matrices). If the matrix is rank deficient, it usually arises from some special structure of the problem (i.e. you usually want to look at where A came from to help you figure out why it is rank deficient, rather than computing the rank etcetera by mindless calculation). (A separate problem is that of matrices that are nearly rank deficient because the pivots are very small, but the right tools to analyze this case won't come up until near the end of the course.)

Further reading: Textbook sections 3.3–3.4, lecture 9.

Lecture 10 (Sep 30)

Dot products, transposes, & orthogonality

Reviewed the dot product or inner product of two real column vectors, x⋅y, defined as ∑ᵢxᵢyᵢ. In linear-algebra terms, we write this as x⋅y=xᵀy in terms of the transpose of the vector x: if x is a column vector, xᵀ is a row vector (sometimes more technically called a "dual" vector or "covector"). The length (or norm) of a vector is is the square root of the dot product with itself: ‖x‖=√xᵀx.

The transpose Aᵀ of a linear operator A is defined so that xᵀAy = x⋅(Ay) = (Aᵀx)⋅y = (Aᵀx)ᵀy: transposes move linear operators from one side to the other in an inner product. In consequence, we find for matrices that Aᵀ is constructed by swapping rows with columns in A. Key properties:

  • (AB)ᵀ=BᵀAᵀ
  • αᵀ = α for scalars. Hence xᵀy = yᵀx (i.e. x⋅y = y⋅x)
  • (A⁻¹)ᵀ = (Aᵀ)⁻¹

Orthogonal vectors are those for which xᵀy = 0. Defined the orthogonal complement S⟂ of a subspace S ⊆ V as {x ∈ V such that xᵀy=0 for all y ∈ S}. Combining a basis for S and S⟂ gives a basis for the whole vector space V, so the dimensions of S and S⟂ sum to the dimension of V.

Taking the orthogonal complements of C(A) and N(A) leads us to the four fundamental subspaces for an m×n matrix A of rank r:

  • column space C(A) ⊆ ℝᵐ, dimension r
  • C(A)⟂ = left nullspace N(Aᵀ) ⊆ ℝᵐ, dimension m-r
  • nullspace N(A) ⊆ ℝⁿ, dimension n-r
  • N(A)⟂ = row space C(Aᵀ) ⊆ ℝⁿ, dimension r

(The consequence of this is an amazing fact: A and Aᵀ have the same rank r, since C(A) and C(Aᵀ) must have the same dimension r. Thus, if you do Gaussian elimination on A and Gaussian elimination on Aᵀ, you will get the same number of pivots in both cases even though the elimination processes are quite different.)

Further reading: Textbook sections 3.5, 4.1; video lecture 10, video lecture 14. Julia notebook on transposes and orthogonality.

Exam 1 (Friday Oct 7 at 11am in 26-100)

Exam 1 will cover the material through lecture 10 and pset 4, including: linear operators, matrix–matrix and matrix–vector operations and interpretations thereof, writing/working with equations in matrix form, solving systems of equations with one or more right-hand sides, Gaussian elimination, back/forward-substitution and triangular matrices, LU factorization and PA=LU, permutation matrices, matrix inverses and Gauss–Jordan, singular matrices, computational costs (which operations are ~ m² or ~ m³ etc and arranging calculations efficiently), rank of a matrix (= number of pivots), vector space & subspaces, null space & special solutions, pivot/free columns column spaces, bases and dimensions of vector spaces, checking whether Ax=b is solvabe, complete solutions to Ax=b (including non-square matrices), transposes and dot products, orthogonality and orthogonal complements, the four fundamental subspaces.

The exam is closed book/notes. (No calculators or computers either.)

  • (Optional) review session: Thursday 5/6 4–5pm via Zoom. A recording and list of practice problems will be posted afterwards.

Some practice problems: spring 2017 exam 1 problems 1–4 (solutions); fall 2017 exam 1 problems 1-4 (solutions); fall 2017 exam 1 problem 1a,b (solutions); fall 2014 exam 1 problems 1, 2, 3, 4c/d/e (solutions); fall 2012 exam 1 problems 1–4 (solutions); spring 2008 exam 1 problems 1–4 (solutions); fall 2011 exam 1 problem 1–3 (solutions); spring 2012 exam 1 problems 1–4 (solutions); fall 2013 exam 1 problems 1, 2, 3, 5 (solutions); fall 2009 exam 1 problems 1, 2, 3, 4 (solutions); spring 2008 exam 1 problems 1, 3, 4 (solutions)

About

18.06 course at MIT

Resources

Readme
Activity

Stars

0 stars

Watchers

0 watching

Forks

0 forks
Report repository

Releases

No releases published

Packages

No packages published

Languages

  • Jupyter Notebook 100.0%
0