The Inverse of an Orthogonal Matrix is its Transpose.

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. 

So Close and Yet So Far: What you can do if you know ALMOST every Eigenvalue.

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?

QR-Factorization.

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 . 

Neat Questions! Invertible Power of a Matrix.

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.

Application of the Real Spectral Theorem.

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 !

The Spectral Theorem, part 2: The Real Part!

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.

The Spectral Theorem, part 1: Complex Version.

(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?

Adjoints!, or: Why Aren’t These in My Crappy Linear Algebra Book? part 2.

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.

Adjoints: A Detailed Example of the Conjugate Transpose Property.

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.

Adjoints!, or: Why Aren’t These in My Crappy Linear Algebra Book? part 1.

Now, for this post, we’re going to assume something kind of unique: namely, we’re going to assume you know how to work inner products.  Yes, alas, I’m going to make a leap of faith here — but, reader, do not let me down!  I’ve posted a pdf explaining what they are a few posts ago, and we’ll be going over some basic properties of them as I go on in this post.  But the best way to learn these is to really do a bunch of problems that deal with them.  Almost every linear algebra book that I’ve seen has a huge section on these.

On the other hand, nearly every (basic) linear algebra book that I’ve seen has either a passing mention of adjoints or no mention at all.  This is more of an advanced topic, but it really doesn’t need to be — it’s not difficult, it’s just not all that intuitive.

But let’s stop talking about it and let’s start doing it.