# The Unapologetic Mathematician

## 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 | Algebra, Linear Algebra | 14 Comments

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 | 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 | Algebra, Linear Algebra | 5 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 | Algebra, Linear Algebra | 12 Comments

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 | Algebra, Linear Algebra | 9 Comments

## 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 | Algebra, Linear Algebra | 2 Comments

## Hiatus

This week’s been hectic, what with interviews (which I can’t talk about but I think went okay) and driving back to Kentucky. And now I’m about to spend two weeks up at Mammoth Cave. I should have some internet access, so I’ll try to keep up with lower-content features, but real mathy stuff is off the table for a while. You can always follow my twitter stream to keep tabs, and find links to pictures I take underground, assuming that the wireless at Hamilton Valley really works. But of course, the best laid plans of mice and men…