August 7, 2012
An interesting question came up during my studying today which made me think about some of the different ways we can think about polynomials. The solution to this problem was not immediately obvious to me, and, in fact, it wasn’t until I looked up a completely unrelated problem (in a numerical methods book!) that some solution became clear.
Question: Suppose that is a sequence of polynomials with each of degree and suppose pointwise. Show that is also a polynomial of degree no more than .
Some interesting points come up here. First is that we only have pointwise convergence — it wasn’t even immediately obvious to me how to prove the resulting limit was continuous, let alone a polynomial of some degree. Second, we know very little about the polynomials except for what degree they are. This should be an indication that we need to characterize them with respect to something degree-related.
Indeed, polynomials can be represented in a few nice ways. Among these are:
- In the form where it is usually stated that .
- In terms of their coefficients. That is, if we have a list of polynomials of degree 3 to store on a computer, we could create an array where the first column is the constant, the second is the linear term, and so forth. This is sort of like decimal expansion.
- Where they send each point. That is, if we know what is equal to for each , we could recover it.
- If a polynomial is of degree then, somewhat surprisingly, we can improve upon the previous statement: if we know the value of for distinct points, then we can find , and is the unique such polynomial of that degree which has those values. (Note that if we were to have points and a polynomial of degree , then many polynomials of this degree could fit the points. Consider, for example, and . Then we have one points and we want to fit line through it. Clearly this can be done in infinitely many ways.)
This last one is going to be useful for us. So much so that it might be a good idea to prove it.
Lemma. Let be distinct points in , and let be distinct points in . Then there is a unique polynomial of degree at most such that for each considered here.
Proof. This is an exercise in linear algebra. We need to solve the system of linear equations
where spans , for the constants . Notice that this is simply plugging into a general polynomial of degree . Notice that the matrix that this forms will be a Vandermonde matrix. Since each is distinct, the determinant of this matrix is nonzero, which implies that there is a unique solution. This gives us our coefficients, and note that this is a polynomial not necessarily of degree exactly , since some coefficients may be 0, but it is at most .
[Note: For those of you who forgot your linear algebra, the end of this goes like this: if we let our coefficients be denoted by the column matrix and our Vandermonde matrix is denoted by , then we want to solve where is the column vector with entries . If has non-zero determinant, then it is invertible, and so we have that gives us our coefficients.]
Neato. But now we need to specialize this somewhat for our proof.
Corollary. Let the notation and assumptions be as in the last lemma. For , let be the unique polynomial of degree at most with (where if and ). Then every polynomial of degree at most is of the form for each .
This might be a bit more cryptic, so let’s do an example. Let’s let so that we have two points. Let’s say and . Then we have is the unique polynomial of degree at most 1 such that and . Of course, this function will be . Now and ; this gives us that . The theorem now states that any polynomial of degree at most can be written in the form
For example, let . Then the lemma says , as we’d expect. The power of this lemma will become clear when we use this in the solution. The proof of this corollary is just a specialization of the previous lemma, so we exclude it.
Solution. Recall, just for notation, that our sequence pointwise. Let’s let be our distinct points, as usual. In addition, let’s let be defined as in the corollary above. Represent each as follows:
for each . Here comes the magic: let and note that at every point, so, in particular, on each and on . We obtain
for each . But this is the sum of polynomials of degrees at most , which gives us that is itself a polynomial of degree at most .
I’ll admit, I did a bit of digging around after finding the lemma above; in particular, this corollary representation of polynomials seems to be a nice way to represent a polynomial if we do not know the coefficients but do know the values at a certain number of points and have that its degree is bounded below that number of points.
Exercise: Try to write out this representation for and . If you’re a programmer, why not make a program that allows you to input some points and some values and spits out the polynomial from the corollary above?
November 30, 2010
This is a really sweet deal. In particular, if we know that our matrix is orthogonal, we can cut down on time finding the inverse significantly. Combined with the spectral theorem (which states that if the matrix is symmetric, there is an orthogonal matrix such that is a diagonal matrix with entries the eigenvalues) this gives us a tool for finding diagnalizations of matrices.
Linear Algebra is strange. On the surface, we have a ton of tricks that we can apply to things to make calculations nicer (diagonalizing matrices, finding orthonormal bases,…) but deeper down a lot of things connect to one-another in really unexpected ways — to me, anyway!
Here’s the problem. We have an matrix called , and we have eigenvalues. We have ALMOST every eigenvalue, but we’re just missing one. What can we do about this?
October 28, 2010
While I was studying for a linear algebra exam, I discovered a deep-seeded love for QR-Factorization. I’m not going to explain why this is important, or why we should care about such a factorization; in fact, I have no idea why this is important or why I should care about this. I was directed to this, so I’ll direct you there too.
Anyhow, here’s the game. Given some square matrix (this is not necessarily, but it makes it easier) , we want to decompose into an orthogonal matrix and an upper-triangular matrix .
September 25, 2010
On a linear algebra exam, I was asked something like, “If is a real matrix and is invertible, prove that is also invertible.”
Apparently, a number of students attempted the problem this way:
and concluded that had an inverse. What’s wrong with this argument? The astute reader will note that we’re actually assuming that exists to prove its existence! That’s a little circular. Instead, we’re going to prove something slightly more general.
August 7, 2010
I mean, okay, we wrote about the real spectral theorem, but how much do we really know about it? A lot, I hope! I hope you remember that we needed to be self-adjoint in order to imply that we have an orthonormal basis of eigenvectors with respect to !
August 1, 2010
Okay, so, last time we talked about the spectral theorem for complex vector spaces. What did it say? Do you remember? Don’t look. Fine, look. Either way, it said that we have an orthonormal basis made of eigenvectors of some linear map if and only if is normal. Now, being normal is not that big of’a deal. It just means that . Not a biggie, right? Yeah.
July 29, 2010
(Note) So, the general spectral theorem is pretty sweet, but (as Sheldon Axler does in Linear Algebra Done Right, the book that I’m essentially following in this blog) I’m going to split it up into two parts. In “real” math, I suppose we should consider two cases: when the field is algebraically closed and when it is not. The algebraically closed case is going to be nearly identical to the complex case. But because we don’t know “how” algebraically closed the other field is, I’m not entirely certain that the “not algebraically closed” case follows from the Reals case of the theorem. For example, if we were to use the integers in place of the reals, we would most likely be able to produce examples which did not follow the Reals version of the spectral proof. Either way…we will mostly be using this “in real life” in the case that the field is either the reals or the complexes. Thus, I do not feel too bad for not proving this in its full generality.
So, let’s wonder something for a second: why have I been proving all these random things? What the hell were we looking for again?
This part is going to be really exciting, no lies. In fact, we’re really just going to prove one or two theorems and that’ll be that. The reason for doing so is because one of these theorems is so elegant and beautiful that I want you to focus on it. Specifically, the fact that the matrix associated to an adjoint linear map is the conjugate transpose of the matrix associated to the original linear map. Best.
My next post is going to be this nice proof of the fact that if we have for some linear map for is a nontrivial finite vector space, then is really easy to find: it’s simply the conjugate transpose of . This is exactly what it sounds like: we find the transpose of (by switching with for all and ) and then replacing every element in the matrix with its conjugate. I guess the notation should be . Which is pretty complicated looking, but it’s really not hard to do at all. In fact, the point of this post is to give a detailed example. Actually, let’s do two examples, and you’re going to find that half of the time this is much easier to do than what I’ve said above.