The Unapologetic Mathematician

Mathematics for the interested outsider

Sunday Samples 128

Gin Blossoms’ major-label breakthrough, New Miserable Experience came out in 1992, while I was still in middle school. There were a whole group of us who really liked the album, but I don’t think any of us realized why until much later on. It leads off with one of the few songs on the album not to be released as a single: “Lost Horizons”.
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July 5, 2009 Posted by John Armstrong | Sunday Samples | | 1 Comment

Dirac Notation II

We continue discussing Dirac notation by bringing up the inner product. To this point, our notation applies to any vector space and its dual, with the ket \lvert v\rangle denoting a vector v\in V and the bra \langle\lambda\rvert denoting a linear functional \lambda\in V^*. The evaluation \lambda(v) is then denoted by the bra-ket pairing \langle\lambda\vert v\rangle.

But the neat thing about this notation is that it makes bras look like some sort of reflection of kets. And they are, in a sense. The dual space V^* is some sort of reflection of the vector space V, but there’s no clear mapping from vectors in one space to vectors in the other; unless, that is, we pick a specific isomorphism; or, equivalently, an inner product.

When we’ve got an inner product in the picture, we get a (conjugate) linear isomorphism that sends the vector v to the linear functional \langle v,\underline{\hphantom{X}}\rangle. In Dirac notation, we send the ket \lvert v\rangle to the bra \langle v\rvert. Then the value of this linear functional on a vector w (the ket \lvert w\rangle) is the pairing \langle v\vert w\rangle=\langle v,w\rangle, just as it should be.

July 1, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 4 Comments

Dirac notation I

There’s a really neat notation for inner product spaces invented by Paul Dirac that physicists use all the time. It really brings to the fore the way a both slots of the inner product enter on an equal footing.

First, we have a bracket, which brings together two vectors

\displaystyle\langle w,v\rangle

the two sides of the product are almost the same, except that the first slot is antilinear — it takes the complex conjugate of scalar multiples — while the second one is linear. Still, we’ve got one antilinear vector variable, and one linear vector variable, and when we bring them together we get a scalar. The first change we’ll make is just to tweak that comma a bit

\displaystyle\langle w\vert v\rangle

Now it doesn’t look as much like a list of variables, but it suggests we pry this bracket apart at the seam

\displaystyle\langle w\rvert\lvert v\rangle

We’ve broken up the bracket into a “bra-ket”, composed of a “ket” vector \lvert v\rangle and a “bra” dual vector \langle w\rvert (pause here to let the giggling subside) (seriously, I taught this to middle- and high-schoolers once).

In this notation, we write vectors in V as kets, with some signifier inside the ket symbol. Often this might be the name of the vector, as in \lvert v\rangle, but it can be anything that sufficiently identifies the vector. One common choice is to specify a basis that we would usually write \left\{e_i\right\}. But the index is sufficient to identify a basis vector, so we might write \lvert1\rangle, \lvert2\rangle, \lvert3\rangle, and so on to denote basis vectors. That is, \lvert i\rangle=e_i. We can even extend this idea into tensor products as follows

\displaystyle e_i\otimes e_j=\lvert i\rangle\otimes\lvert j\rangle=\lvert i,j\rangle

Just put a list of indices inside the ket, and read it as the tensor product of a list of basis vectors.

Bras work the same way — put anything inside them you want (all right, class…) as long as it specifies a vector. The difference is that the bra \langle w\rvert denotes a vector in the dual space V^*. For example, given a basis for V, we may write \langle i\rvert=\epsilon^i for a dual basis vector.

Putting a bra and a ket together means the same as evaluating the linear functional specified by the bra at the vector specified by the ket. Or we could remember that we can consider any vector in V to be a linear functional on V^*, and read the bra-ket as an evaluation that way. The nice part about Dirac notation is that it doesn’t really privilege either viewpoint — both the bra and the ket enter on an equal footing.

June 30, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 12 Comments

Sunday Samples 127 (one day late)

Okay, everybody in the world now knows that Michael Jackson died. And yes, it’s a big deal. But there’s a lot of reputation-burnishing going on out there, one piece of which I’d like to talk about today.

Everyone talks about the music video for “Thriller”, which was really great. It was practically a movie in its own right. But just as Rob Reiner can make North and Steven Spielberg can make The Lost World, so could Michael Jackson release a great, big stinker. And the one that comes most readily to mind is 1991’s “Black or White”. Yes, it hit number one — and fast — but at its heart it’s a pop-rock tune that by turns panders to the current vogues of dance and rap music. It’s got a decent message, but one which has gotten better (and bolder) treatments elsewhere.

But all that is beside the point. The full impact doesn’t set in until you look at the music video. What usually got aired (and what is being aired most places I see it now) is actually pretty good. In particular, it’s very well-known for the sequence using some of the earliest video-quality morphing software to transform a sequence of faces of varied race and sex into each other. Neat. But then there’s the framing.

The full version of the video opens with George Wendt yelling at Macaulay Culkin (fresh from Home Alone) over playing his rock music too loud, and backed up by Tess Harper playing a dishrag of a mother. The music Culkin plays has nothing to do with the actual song, nor does the stereotyped argument. It serves only to cartoonishly launch Wendt (and his armchair) from suburban USA to sound-stage Africa, where the music video proper begins, and Wendt is promptly forgotten. Two minutes that add precisely zero, except that Culkin (whose own presence is a form of pandering) shows up as a caricature in one of the rap interludes.

Okay, fine. “Thriller” had a framing story too. It was more integrated into its video, and made some remote sense, but this much is not entirely without precedent. But then there’s the ending. After the morphing sequence we cut to an overhead shot of the studio and overhear the director ask the last woman from the video “how do you do that?” as if nobody had ever heard of visual effects by 1991. Then we move over to see a panther — a panther — which had nothing whatsoever to do with the video slink off set, where it turns into Michael Jackson. For the next four and a half minutes — as long as the video itself was — Jackson dances around a sound-stage street with no music (but with foley-enhanced sound effects) smashing things and grabbing himself. If nothing else, this is what cemented his reputation for that style of dancing. Yes, some of the stuff he smashes has perfunctory divisive language on it, and there’s some vain attempt at symbology, but the whole segment boils down to “watch me dance”. And dancing without music gets really boring after a while. Definitely before four minutes is up.
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June 29, 2009 Posted by John Armstrong | Sunday Samples | | 5 Comments

Updates

As a much-anticipated visit approaches, my apartment is asymptotically approaching neatness. My former students’ comprehension of the word “asymptotically”, however, remains constant.

And this morning I received the proofs to a paper that was accepted for publication two years ago: “Functors Extending the Kauffman Bracket”. So yeah, I get out with it finally appearing in print, but in the meantime my career is shot.

June 26, 2009 Posted by John Armstrong | Uncategorized | | 7 Comments

Matrices and Forms II

Let’s pick up our discussion of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form B:V\otimes V\rightarrow\mathbb{F}$ on a vector space V over the real or complex numbers, which we can also think of as a linear map from V to its dual space V^*. We can also consider a sesquilinear form on a complex vector space, which is equivalent to an antilinear (conjugates scalars) map from the space to its dual.

We got a matrix by picking a basis \left\{e_i\right\} for V and plugging basis vectors into our form

\displaystyle b_{ij}=B(e_i,e_j)

And we also found that this is the matrix of the map from V to V^*, written in terms of the basis \left\{e_i\right\} and its dual basis \left\{\epsilon^i\right\}.

Now we have two bases of the same cardinality, so there is a (highly non-canonical) linear isomorphism from V^* to V which sends the basis vector \epsilon^i to the basis vector e_i. If we compose this with the map from V to V^* given by the form B, we get a linear map from V to itself, which we will also call B. If B is sesquilinear, we use the unique antilinear isomorphism sending \epsilon^i to e_i, and again get a linear map B:V\rightarrow V. The matrix of this linear map with respect to the basis \left\{e_i\right\} is b_i{}^j=b_{ij}, just as before.

This seems sort of artificial, but there’s method here. Remember that we can come up with an inner product \langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle by simply declaring the basis \left\{e_i\right\} to be orthonormal. Then the linear functional \epsilon^i is given by \langle e_i,\underline{\hphantom{X}}\rangle. It’s this correspondence that is captured by both the non-canonical choice of inner product, and by the non-canonical choice of isomorphism.

But now we can extract the same matrix in a slightly different way. We start by hitting the basis vector e_j with the new linear transformation B to come up with a linear combination of the basis vectors. We then extract the appropriate component by using the inner product. In the end, we find

\displaystyle\langle e_i,B(e_j)\rangle=b_{ij}=B(e_i,e_j)

So in the presence of a given basis (or at least a given inner product) any bilinear (or sesquilinear) form corresponds to a linear transformation, and the matrix of the linear transformation with respect to the selected basis is exactly that of the form itself. On the other hand, if we have a linear transformation from V to itself, we can stick it into the inner product as above and get a bilinear (or sesquilinear) form out.

June 25, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 2 Comments

Matrices and Forms I

Yesterday, we defined a Hermitian matrix to be the matrix-theoretic analogue of a self-adjoint transformation. So why should we separate out the two concepts? Well, it turns out that there are more things we can do with a matrix than represent a linear transformation. In fact, we can use matrices to represent forms, as follows.

Let’s start with either a bilinear or a sesquilinear form B\left(\underline{\hphantom{X}},\underline{\hphantom{X}}\right) on the vector space V. Let’s also pick an arbitrary basis \left\{e_i\right\} of V. I want to emphasize that this basis is arbitrary, since recently we’ve been accustomed to automatically picking orthonormal bases. But notice that I’m not assuming that our form is even an inner product to begin with.

Now we can define a matrix b_{ij}=B(e_i,e_j). This completely specifies the form, by either bilinearity or sesquilinearity. And properties of such forms are reflected in their matrices.

For example, suppose that H is a conjugate-symmetric sesquilinear form. That is, H(v,w)=\overline{H(w,v)}. Then we look at the matrix and find

\displaystyle\begin{aligned}h_{ij}&=H\left(e_i,e_j\right)\\&=\overline{H\left(e_j,e_i\right)}\\&=\overline{h_{ji}}\end{aligned}

so H is a Hermitian matrix!

Now the secret here is that the matrix of a form secretly is the matrix of a linear transformation. It’s the transformation that takes us from V to V^* by acting on one slot of the form, and written in terms of the basis e_i and its dual. Let me be a little more explicit.

When we feed a basis vector into our form B, we get a linear functional B(e_i,\underline{\hphantom{X}}). We want to write that out in terms of the dual basis \left\{\epsilon^j\right\} as a linear combination

\displaystyle B(e_i,\underline{\hphantom{X}})=b_{ik}\epsilon^k

So how do we read off these coefficients? Stick another basis vector into the form!

\displaystyle\begin{aligned}B(e_i,e_j)&=b_{ik}\epsilon^k(e_j)\\&=b_{ik}\delta^k_j\\&=b_{ij}\end{aligned}

which is just the same matrix as we found before.

June 24, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 8 Comments

I Made It!

Tonight, in an effort to use up some frozen meat before I move out, I made

Pork Chops with Mango Pineapple Sauce

Pork Chops with Mango Pineapple Sauce

“Pork Chops with Mango Pineapple Sauce”. Some comments:

First, if you can’t find mango nectar, apricot nectar is more readily available and should substitute.

Second, get the pineapple chunks packed in pineapple juice rather than syrup. And guys, don’t just drain off the juice; drink it. Trust me.

Third, mangoes are notable for being some of the most difficult fruit to work with. But don’t let that stop you here, even if you don’t have a special, single-purpose tool for seeding and chopping a mango. Go ahead and just get messy and mangle the flesh as you peel and seed the fruit because you’re going to purée it anyway.

June 24, 2009 Posted by John Armstrong | I Made It! | | No Comments Yet

Self-Adjoint Transformations

Let’s now consider a single inner-product space V and a linear transformation T:V\rightarrow V. Its adjoint is another linear transformation T^*:V\rightarrow V. This opens up the possibility that T^* might be the same transformation as T. If this happens, we say that T is “self-adjoint”. It then satisfies the adjoint relation

\displaystyle\langle v,T(w)\rangle=\langle T(v),w\rangle

What does this look like in terms of matrices? Since we only have one vector space we only need to pick one orthonormal basis \left\{e_i\right\}. Then we get a matrix

\displaystyle\begin{aligned}t_i^j&=\langle e_j,T(e_i)\rangle\\&=\langle T(e_j),e_i\rangle\\&=\overline{\langle e_i,T(e_j)\rangle}\\&=\overline{t_j^i}\end{aligned}

That is, the matrix of a self-adjoint transformation is its own conjugate transpose. We have a special name for this sort of matrix — “Hermitian” — even though it’s exactly equivalent to self-adjointness as a linear transformation. If we’re just working over a real vector space we don’t have to bother with conjugation. In that case we just say that the matrix is symmetric.

Over a one-dimensional complex vector space, the matrix of a linear transformation T is simply a single complex number t. If T is to be self-adjoint, we must have t=\bar{t}, and so t must be a real number. In this sense, the operation of taking the conjugate transpose of a complex matrix (or the simple transpose of a real matrix) extends the idea of conjugating a complex number. Self-adjoint matrices, then, are analogous to real numbers.

June 23, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 1 Comment

The Matrix of the Adjoint

I hear joints popping as I stretch and try to get back into the main line of my posts.

We left off defining what we mean by a matrix element of a linear transformation. Let’s see how this relates to adjoints.

We start with a linear transformation T:V\rightarrow W between two inner product spaces. Given any vectors v\in V and w\in W we have the matrix element \langle w,T(v)\rangle_W, using the inner product on W. We can also write down the adjoint transformation T^*:W\rightarrow V, and its matrix element \langle v,T^*(w)\rangle_V, using the inner product on V.

But the inner product on W is (conjugate) linear. That is, we know that the matrix element \langle w,T(v)\rangle_W can also be written as \overline{\langle T(v),w\rangle_W}. And we also have the adjoint relation \langle v,T^*(w)\rangle_V=\langle T(v),w\rangle_W. Putting these together, we find

\displaystyle\begin{aligned}\langle v,T^*(w)\rangle_V&=\langle T(v),w\rangle_W\\&=\overline{\langle w,T(v)\rangle_W}\end{aligned}

So the matrix elements of T and T^* are pretty closely related.

What if we pick whole orthonormal bases \left\{e_i\right\} of V and \left\{f_j\right\} of W? Now we can write out an entire matrix of T as t_i^j=\langle f_j,T(e_i)\rangle_W. Similarly, we can write a matrix of T^* as

\displaystyle\begin{aligned}\left(t^*\right)_j^i&=\langle e_i,T^*(f_j)\rangle_V\\&=\overline{\langle f_j,T(e_i)\rangle_W}\\&=\overline{t_i^j}\end{aligned}

That is, we get the matrix for the adjoint transformation by taking the original matrix, swapping the two indices, and taking the complex conjugate of each entry. This “conjugate transpose” operation on matrices reflects adjunction on transformations.

June 22, 2009 Posted by John Armstrong | Algebra, Linear Algebra | | 1 Comment

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