This is archived information for Math 221 Sect 101 (Fall, 2002).
Here's a summary of what I remember talking about for the review session last night (Friday, December 13).
First, I talked for about 20 minutes just on some general advice for studying for this exam. Here are my slides.
It looks like, over the course of the term, we studied roughly 100 definitions and 80 theorems and other miscellaneous facts. Relearning them in detail over the weekend would be difficult. However, it's worth looking at the important terms, the things like "basis", "linear independence", and so on that kept coming up again and again. When you think of "basis", its definition---that it's a linearly independent set that spans a space---should immediately pop into your head. And dancing around that definition should be a little parade of related concepts: the Spanning Set Theorem, the Basis Theorem, how to find a basis for a column space or null space, and so on.
Besides these important terms and related concepts, there were several hefty algorithms and procedures that we used, everything from doing row reduction to solve a linear system up to all the steps involved in doing an orthogonal diagonalization of a symmetric matrix. The best references for these are the quizzes and homework. The last slide in the file above gives a quick list I made just by looking through the quizzes and listing any significant calculations I could find. It's not a complete list (I probably missed some things, and I know there were a few things assigned as homework that weren't on any quizzes), but it should get you started.
Next, someone asked me to clarify what exactly a "distinct" eigenvalue is supposed to be. Is it something different from a regular eigenvalue? Well, no. They're all the same eigenvalues. It just depends on how they're counted.
This talk about "distinct" eigenvalues arises because of the concept of the multiplicity of an eigenvalue. The characteristic equation for a matrix gives all of its eigenvalues. For example, a 3x3 matrix might have a characteristic equation:
det(A-lambda I) = (lambda-6)(lambda-5)(lambda-6) = 0
If we count multiplicities, this matrix has three eigenvalues: the eigenvalue 5 has multiplicity 1, and the eigenvalue 6 has multiplicity 2 because it appears in 2 factors of the polynomial. The matrix has only two distinct eigenvalues though---only two eigenvalues with different values.
The concept of multiplicity (allowing the same distinct eigenvalue to be counted multiple times if its factor appears multiple times in the characteristic polynomial) makes the statement of many theorems much simpler. For example, an nxn matrix always has n (possibly complex-valued) eigenvalues counting multiplicity. The number of distinct eigenvalues might be smaller, however.
In other contexts, it's important to emphasize that we're talking about eigenvalues with different values. For example, we have a theorem which says that an nxn matrix with n distinct eigenvalues is diagonalizable. Every matrix has n eigenvalues counting multiplicities. But those that have n distinct eigenvalues are automatically diagonalizable.
Next, I believe there were a couple of questions about coordinate vectors and B-matrices of transformations, things that popped up in Sections 4.4 and 5.4. In particular, I was asked to explain Figure 3 on page 323.
Remember long, long ago we studied the Unique Representation Theorem. This told us that, if B={b1,...,bn} is a basis for Rn, then every vector in Rn can be represented as a linear combination of the vectors of B and this representation is unique in the sense that only one set of weights will work. This motivates a definition. For a fixed basis B, we can define [x]B to be the (unique) vector of weights (c1,...,cn) that satisfies
x=c1b1+...+cnbn
That is, every vector in Rn can be represented instead by its coordinates (its weights) with respect to the basis B. Graphically, this is like laying down some funny graph, ignoring the actual locations of the vectors and, instead, measuring them all with respect to the funny graph paper. There's a pretty good explanation in Section 4.4. See the figures at the top of page 242 and the associated explanation on the previous page.
Okay, back to that diagram on page 323. In this diagram, T is a linear transformation from a vector space to itself. For our purposes, let's say T is a transformation from Rn to Rn (and so B is a basis for the same space Rn). The procedure above, measuring vectors using funny graph paper, accounts for the downward-pointing arrows on the left and right. On the left, we're taking any old vector x, and we're calculating its funny-graph-paper-coordinates, [x]B. On the right, we're starting with the image of x under the transformation T. But, this is just some other vector in Rn, so there's no reason we can't look at its funny-graph-paper-coordinates [T(x)]B, too. In addition, the top arrow is self-explanatory. If we take the vector x and apply the linear transformation T (which amounts to multiplying x by T's standard matrix A, really), we get T(x). The only mysterious arrow is the bottom one.
What the bottom arrow indicates is that, to someone who ignores the actual locations of the vectors and sees only their coordinates on the funny graph paper, the linear transformation T, which is actually multiplication by T's standard matrix A, looks like a different linear transformation, which respect to the funny graph paper. This new transformation has standard matrix [T]B. That is, to a "normal" person, the transformation looks like multiplication by a matrix A. To a person wearing funny glasses that makes the funny graph paper look like normal graph paper, the transformation looks like multiplication by a different matrix [T]B.
Practically speaking, if you need to calculate a B-matrix [T]B, it's just a nasty twist on the calculation of the standard matrix. To calculate a standard matrix for T, you create a matrix whose columns are:
A=[T(e1) T(e2) ... T(en)]
To calculate a B-matrix for T, you create a matrix whose columns are:
A=[ [T(b1)]B [T(b2)]B ... T(bn)]B ]
For each column, there are two funny things going on. First, instead of calculating the T-image of a standard basis vector like e1, we calculate the T-image of one of the B-basis vectors b1. Second, we don't use that result as-is. Instead, we take the further step of calculating the B-coordinates of the result, using [T(b1)]B instead of using T(b1) directly.
I was also asked to go over Midterm #1, Question #5 again. The solution is still available in the key to Midterm #1, so I won't repeat it here.
There were a few more questions, I think, but I can't remember what they were.
This is archived information for Math 221 Sect 101 (Fall, 2002).
Last revised: Wed Jan 29 12:48:08 PST 2003