The Unapologetic Mathematician

Mathematics for the interested outsider

Matrix Elements

Okay, back to linear algebra and inner product spaces. I want to look at the matrix of a linear map between finite-dimensional inner product spaces.

So, let’s say V and W are inner product spaces with orthonormal bases \left\{e_i\right\}_{i=1}^m and \left\{f_j\right\}_{j=1}^n, respectively, and let T:V\rightarrow W be a linear map from one to the other. We know that we can write down the matrix \left(t_i^j\right) for T, where the matrix entries are defined as the coefficients in the expansion

\displaystyle T(e_i)=t_i^kf_k

But now that we’ve got an inner product on W, it will be easy to extract these coefficients. Just consider the inner product

\displaystyle\begin{aligned}\left\langle f_j,T(e_i)\right\rangle&=\left\langle f_j,t_i^kf_k\right\rangle\\&=t_i^k\left\langle f_j,f_k\right\rangle\\&=t_i^k\delta_{jk}\\&=t_i^j\end{aligned}

Presto! We have a nice, neat function that takes a linear map T and gives us back the i-j entry in its matrix — with respect to the appropriate bases, naturally.

But this is also the root of a subtle, but important, shift in understanding what a matrix entry actually is. Up until now, we’ve thought of matrix entries as artifacts which happen to be useful for calculations. But now we’re very explicitly looking at the question “what scalar shows up in this slot of the matrix of a linear map with respect to these particular bases?” as a function. In fact, t_i^j is now not just some scalar value peculiar to the transformation at hand; it’s now a particular linear functional on the space of all transformations \hom(V,W).

And, really, what do the indices i and j matter? If we rearranged the bases we’d find the same function in a new place in the new array. We could have taken this perspective before, with any vector space, but what we couldn’t have asked before is this more general question: “Given a vector v\in V and a vector w\in W, how much does the image T(v) is made up of w?” This new question only asks about these two particular vectors, and doesn’t care anything about any of the other basis vectors that may (or may not!) be floating around. But in the context of an inner product space, this question has an answer:

\displaystyle\left\langle w,T(v)\right\rangle

Any function of this form we’ll call a “matrix element”. We can use such matrix elements to probe linear transformations T even without full bases to work with, sort of like the way we generalized “elements” of an abelian group to “members” of an object in an abelian category. This is especially useful when we move to the infinite-dimensional context and might find it hard to come up with a proper basis to make a matrix with. Instead, we can work with the collection of all matrix elements and use it in arguments in place of some particular collection of matrix elements which happen to come from particular bases.

Now it would be really neat if matrix elements themselves formed a vector space, but the situation’s sort of like when we constructed tensor products. Matrix elements are like the “pure” tensors v\otimes w\in V\otimes W. They (far more than) span the space \hom(V,W)^* of all linear functionals on the space of linear transformations, just like pure tensors span the whole tensor product space. But almost all linear functionals have to be written as a nontrivial sum of matrix elements — they usually can’t be written with just one. Still, since they span we know that many properties which hold for all matrix elements will immediately hold for all linear functionals on T.

May 29, 2009 Posted by | Algebra, Linear Algebra | 2 Comments

Complex Numbers and Polar Coordinates

Forgot to hit “publish” earlier…

So we’ve seen that the unit complex numbers can be written in the form e^{i\theta} where \theta denotes the (signed) angle between the point on the circle and 1+0i. We’ve also seen that this view behaves particularly nicely with respect to multiplication: multiplying two unit complex numbers just adds their angles. Today I want to extend this viewpoint to the whole complex plane.

If we start with any nonzero complex number z=a+bi, we can find its absolute value \lvert z\rvert=\sqrt{a^2+b^2}. This is a positive real number which we’ll also call r. We can factor this out of z to find z=r\left(\frac{a}{r}+\frac{b}{r}i\right). The complex number in parentheses has unit absolute value, and so we can write it as e^{i\theta} for some \theta between -\pi and \pi. Thus we’ve written our complex number in the form

\displaystyle z=re^{i\theta}

where the positive real number r is the absolute value of z, and \theta — a real number in the range \left(-\pi,\pi\right] — is the angle z makes with the reference point 1+0i. But this is exactly how we define the polar coordinates (r,\theta) back in high school math courses.

Just like we saw for unit complex numbers, this notation is very well behaved with respect to multiplication. Given complex numbers r_1e^{i\theta_1} and r_2e^{i\theta_2} we calculate their product:

\displaystyle r_1e^{i\theta_1}r_2e^{i\theta_2}=\left(r_1r_2\right)e^{i\left(\theta_1+\theta_2\right)}

That is, we multiply their lengths (as we already knew) and add their angles, just like before. This viewpoint also makes division simple:

\displaystyle\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2}e^{i\left(\theta_1-\theta_2\right)}

In particular we see that

\displaystyle\left(re^{i\theta}\right)^{-1}=\frac{1}{r}e^{-i\theta}=\frac{1}{r^2}\overline{re^{i\theta}}

so multiplicative inverses are given in terms of complex conjugates and magnitudes as we already knew.

Powers (including roots) are also easy, which gives rise to easy ways to remember all those messy double- and triple-angle formulæ from trigonometry:

\displaystyle\begin{aligned}\cos(2\theta)+i\sin(2\theta)&=e^{i\left(2\theta\right)}\\&=\left(e^{i\theta}\right)^2\\&=\left(\cos(\theta)+i\sin(\theta)\right)^2\\&=\left(\cos(\theta)^2-\sin(\theta)^2\right)+i\left(2\sin(\theta)\cos(\theta)\right)\end{aligned}

\displaystyle\begin{aligned}\cos(3\theta)+i\sin(3\theta)&=e^{i\left(3\theta\right)}\\&=\left(e^{i\theta}\right)^3\\&=\left(\cos(\theta)+i\sin(\theta)\right)^3\\&=\left(\cos(\theta)^3-3\cos(\theta)\sin(\theta)^2\right)+i\left(3\cos(\theta)^2\sin(\theta)-\sin(\theta)^3\right)\\&=\left(\cos(\theta)^3-3\cos(\theta)\left(1-\cos(\theta)^2\right)\right)+i\left(3\left(1-\sin(\theta)^2\right)\sin(\theta)-\sin(\theta)^3\right)\\&=\left(4\cos(\theta)^3-3\cos(\theta)\right)+i\left(3\sin(\theta)-4\sin(\theta)^3\right)\end{aligned}

Other angle addition formulæ should be similarly easy to verify from this point.

In general, since we consider complex numbers multiplicatively so often it will be convenient to have this polar representation of complex numbers at hand. It will also generalize nicely, as we will see.

May 29, 2009 Posted by | Fundamentals, Numbers | 3 Comments

   

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