34. Final Course Review
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Overview
An MIT linear algebra course review works through exam-style questions connecting matrix rank, null spaces, determinants, projections, eigenvalues, Markov processes, and least squares. The examples include a 3-by-n matrix with rank r below 3 but independent columns, a Markov chain with eigenvalues 0, 1, and -0.2 whose steady state is (3, 3, 4), and a line-fitting least-squares solution of (11/3, -1).
Key takeaways
- A system with at least one unsolvable right-hand side has rank below its number of rows, while a unique solution for a solvable right-hand side requires a trivial null space.
- For any matrix A, AᵀA is invertible exactly when A has independent columns; AAᵀ is positive definite only when A has full row rank.
- The dimension of the null space of Aᵀ is m - r for an m-by-n matrix A, so a full-row-rank Aᵀ system can still have infinitely many solutions when m > r.
- In a Markov process with eigenvalues 0, 1, and -0.2, only the eigenvalue-1 component survives in the long run; total-population conservation determines the scale of the steady-state eigenvector.
- Least squares projects b onto the column space of A; if b is orthogonal to every column, the projection and coefficient vector are both zero.
Chapters
- The matrix has 3 rows because its right-hand sides include (1, 0, 0) and (0, 1, 0).
- No solution for one right-hand side means the rank r is less than 3, while a unique solution for another means the null space contains only zero.
- Together, the conditions give r = n < 3; a column space containing (0, 1, 0) but not (1, 0, 0) provides an example.
- Because the original matrix has independent columns, AᵀA is invertible; its size is n by n.
- AAᵀ is a 3-by-3 matrix of rank r < 3, so it is positive semidefinite but not positive definite.
- The determinants of AᵀA and AAᵀ need not be equal when A is rectangular; the determinant product rule applies to square matrices.
- Aᵀy = c is solvable for every c because Aᵀ has full row rank, and it has infinitely many solutions because its null space has dimension 3 - r > 0.
- For A with columns v₁, v₂, and v₃, the equation Ax = v₁ - v₂ + v₃ has solution x = (1, -1, 1).
- If v₁ - v₂ + v₃ = 0, then (1, -1, 1) lies in the null space, so any solvable system Ax = b cannot have a unique solution.
- When v₁, v₂, and v₃ are orthonormal, the projection of v₃ onto the plane spanned by v₁ and v₂ is zero.
- A column relation, column 1 plus column 2 equals twice column 3, gives a nonzero null vector and eigenvalue 0.
- The Markov matrix has eigenvalue 1, and its trace of 0.8 implies the remaining eigenvalue is -0.2.
- Starting from u₀ = (0, 10, 0), the powers of A retain only the eigenvalue-1 component as k tends to infinity.
- An eigenvector for eigenvalue 1 is (3, 3, 4); conservation of the total population fixes its coefficient at 1, giving steady state (3, 3, 4).
- For projection onto the line through a = (4, -3), use P = aaᵀ/(aᵀa).
- Given eigenpairs (0, (1, 2)) and (3, (2, 1)), construct A as SΛS⁻¹, with the eigenvectors as columns of S.
- Every matrix BᵀB is symmetric, so any nonsymmetric 2-by-2 matrix is an example that cannot have this factorization.
- Skew-symmetric and orthogonal matrices can have orthogonal eigenvectors without being symmetric; the examples discussed can involve complex eigenvectors.
- For A with columns (1, 1, 1) and (0, 1, 2), and b = (3, 4, 1), the least-squares coefficients are (11/3, -1).
- The projected vector is A x̂ = (11/3, 8/3, 5/3), the point in A's column space closest to b.
- Interpreting the rows as measurements at t = 0, 1, and 2 gives a best-fit line with intercept 11/3 and slope -1.
- A right-hand side orthogonal to both columns of A has zero projection onto the column space, so its least-squares solution is zero.
- For columns (1, 1, 1) and (0, 1, 2), b = (1, -2, 1) is nonzero and orthogonal to each column.
- This example links the least-squares solution to the orthogonal complement of the column space.
Transcript excerpt
0:07 OK. Good. The final class in linear algebra at MIT this Fall is to review the whole course. And, you know the best way I know how to review is to take old exams and just think through the problems. So it will be a three-hour exam next Thursday. Nobody will be able to take an exam before Thursday, anybody who needs to take it in some different way after Thursday should see me next Monday. I'll be in my office Monday. OK. May I just read out some problems
0:54 and, let me bring the board down, and let's start. OK. Here's a question. This is about a 3-by-n matrix. And we're given -- so we're given -- given -- A x equals 1 0 0 has no solution. And we're also given A x equals 0 1 0 has exactly one solution. OK.
1:48 So you can probably anticipate my first question, what can you tell me about m? It's an m-by-n matrix of rank r, as always, what can you tell me about those three numbers? So what can you tell me about m, the number of rows, n, the number of columns, and r, the rank? OK. See, do you want to tell me first what m is? How many rows in this matrix? Must be three, right? We can't tell what n is, but we can certainly tell that m is three.
2:43 OK. And, what do these things tell us? Let's take them one at a time. When I discover that some equation has no solution, that there's some right-hand side with no answer, what does that tell me about the rank of the matrix? It's smaller m. Is that right? If there is no solution, that tells me that some rows of the matrix are combinations of other rows.
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