Showing posts with label eigenvalues. Show all posts
Showing posts with label eigenvalues. Show all posts

From polarisation to quantum mechanics: states, observables, Born's law

Like most texts on the theory, I will motivate the mathematics of quantum mechanics from the example of polarisation -- mostly because it's a very accessible example of stuff being wavelike. From this example, we will be able to motivate: the state vector (generalising the polarisation), state vector collapse (the event of polarisation), observables and their eigenvalues (stuff like energy, number of photons, etc.), eigenstates and their orthogonality (polarisation basis), noncommuting operators and uncertainty (the noncommuting of lenses).

The key feature of quantum mechanics -- the fundamentally probabilistic nature -- comes from the following two facts, confirmed by experiments (the famous experiments here are the double-slit experiment and photoelectric effect respectively):

  • Everything is a wave -- objects behave as waves, following the superposition principle and the waves represent densities of observations at large scales.
  • Everything is a particle -- which manifests itself in the form of some stuff, like energy and momentum, coming in little quanta.

This is the principle of wave-particle duality. You may realise how this implies a probabilistic description, but the following example should make it quite clear: consider a wave of light, with energy $hf$ (so it's a single photon) polarised at angle $\theta$ to the horizontal -- and it passes through a horizontal polarising filter. Well, then the wave that passes through would be a horizontally polarised wave with energy $hf\cos^2\theta$, right?


But this is impossible, since energy levels in quantum mechanics are quantised -- you can't have $\cos^2\theta$ of a photon, you can only have integer multiples of a photon. But the fact that energy drops as $E\cos^2\theta$ is something that you can verify at your home, using sunglasses -- what the heck?

The key point is that the empirical verification of the $\cos^2\theta$ business that you can do at home is on a macroscopic level, when you have a large number of photons $E=Nhf$. So something occurs with the photons on a microscopic level such that when you try it with a large number of photons, $\cos^2\theta$ of the photons pass through.

Well, this is essentially the "definition" of probability! A single photon passes through the filter with a probability of $\cos^2\theta$ so that for a large number of photons, $\cos^2\theta$ of the photons pass through. This is a non-trivial result -- wave-particle duality makes no mentions of probability as such, it just tells us that stuff is both a particle and a wave, but this simple condition in itself implies a probabilistic, non-deterministic reality.



Similar thought experiments can illustrate the probabilistic nature of other things (the "things" in question here will soon be called "eigenstates"): position is easy -- consider a standing wave photon in a box (this can easily be constructed). This is uniformly distributed throughout the box -- so how much of the energy is in some chunk of the box?

Momentum is trickier, but shouldn't be too hard if you're familiar with Fourier transforms -- what's the analog of a "box" in momentum-space? Well, consider a concentrated pulse of light -- this can be written, via a Fourier transform, as the sum of several light waves of different momenta (i.e. frequencies), each wave with some lower energy. Taking "some chunk" of this "box" amounts to filtering some specific frequencies of the light. This can be done easily, e.g. with a colour filter -- so how much of the energy is contained in the waves with these specific momenta?

In both cases, the key point is that you can't have a fraction of the energy of the photon at these positions/momenta, so you must have a probability of measuring the photon to be in a specific range of positions or a specific range of positions -- to be in a specific region or in a specific region of momentum-space.



The fundamental point here can be made for any quantity $X$: if you can filter out the "part" of a collection of particles that has $X$ in a certain subset of its range, then on a microscopic level, is probabilistic. The act of "filtering out the parts with a certain $X$", applied to a single particle, is just the act of checking if a particle is in a certain $X$-interval, and is called measurement. Any quantity that you can measure is called an observable. 

Something like polarisation is really a form of measurement -- you're finding out whether or not the photon is in a certain polarisation $|\phi_{\parallel}\rangle$. You may have another observable, corresponding to a different polarisation -- even one that is orthogonal to the first polarisation -- $|\phi_{\perp}\rangle$ and still get that the photon is in $|\phi_\perp\rangle$. There is nothing wrong with this, as we just know beforehand that the photon is in $|\phi_\parallel\rangle$ or $|\phi_\perp\rangle$. If you perform the polarisation with $|\phi_\perp\rangle$ after the polarisation with $|\phi_\parallel\rangle$, you will find that the photon doesn't pass through, as you know for sure that the photon is not in both $|\phi_\parallel\rangle$ and $|\phi_{\perp}\rangle$.

Now, you may have certain psychological issues with this, as have many in history -- however, you might want to note that the aim of quantum mechanics is not to fix your psychological problems but to explain nature. You need to accept logical positivism and learn to shut up and calculate to be comfortable with quantum mechanics.

So whatever calculus we invent to describe these probabilistic phenomena, it is going to apply to all observables.

In our first example, the polarisation of the photon can be represented by a unit vector which we will denote as $|\psi\rangle$. The polarising filter has two special axes, represented by unit vectors $|\phi_{\parallel}\rangle$ and $|\phi_\perp\rangle$ -- these are special in the sense that an incoming photon polarised as $|\phi_{\parallel}\rangle$ or $|\phi_\perp\rangle$ will simply be scaled, by factors of 1 and 0 respectively -- so these form an eigenbasis for a certain operator.

Well, we said that the photon passes through (with polarisation $|\phi_{\parallel}\rangle$) with probability $\cos^2\theta$ -- this arises simply from considering the amplitude of $|\psi\rangle$ in the direction of $|\phi_\parallel\rangle$. So we can write the probability that the photon ends up in a state $|\phi\rangle$ as $|\langle\psi|\phi\rangle|^2$ where $\langle\psi|\phi\rangle$ is called the corresponding "probability amplitude".

This expression, $P(x=\lambda)=|\langle\psi|\phi_\lambda\rangle|^2$ is called Born's rule.

Let's get back to the eigenbasis -- what exactly is this an eigenbasis of? We said that the corresponding eigenvalues are 1 and 0, so this gives us a complete description of the operator. Note that this operator depends only on the observable (namely "number of photons in the $|\phi_{\parallel}\rangle$ direction), not on the state or any other feature of the observation. So we decide to call this operator/matrix the "observable", and its eigenvalues are the values of the observable that can be measured.

To find properties of these observables, the natural way is to note that the only feature we've really required of them is Born's rule, i.e. the probabilistic interpretation -- so we can apply the axioms of probability and see what they apply in the context of these observables.

  • $P(E)\ge 0$ -- imply that the observables are over either the reals or complexes, so that $|\langle\psi|\phi\rangle|^2\in \mathbb{R}$ in the first place. The nonnegativity then follows.
  • $P(\Omega)=1$ and $P\left(\bigcup_i E_i\right) = \sum_i P(E_i)$ for disjoint $E_i$ -- this, along with the second axiom, implies that $\sum |\langle\phi|\psi\rangle|^2 = 1=|\langle\psi|\psi\rangle|^2$ where the sum is taken over all eigenstates $|\phi\rangle$ of the operator. As this must be true for all states $|\psi\rangle$, the thing on the left must be a Pythagorean sum, so the $|\phi\rangle$s must form an orthogonal basis. This implies that all observables are normal operators.

The latter fact is very important, and can also be seen in the following way -- if you a system is in one eigenstate, it cannot possibly collapse onto another eigenstate (the probabilistic interpretation is: if you know for sure the value of the symbol is a thing, it's that thing) -- so we must have $|\langle \phi_1|\phi_2\rangle|^2=0$ for all eigenstates $|\phi_1\rangle$ and $|\phi_2\rangle$.

Another restriction we add is that the observables be not only normal, but Hermitian operators in particular, so they have real eigenvalues. This may seem an odd choice, but it makes sense, as any normal operator may be uniquely written as $X_H+iX_{AH}$ where $X_H$ and $X_{AH}$ are Hermitian, and $X_H$ and $X_{AH}$ commute, so any complex observation can be done unambiguously as two real observations. So we stick to real eigenvalues.

This also makes it essential that we allow complex operators rather than just real ones (the two choices were given to us from the first probability axiom), so that this decomposition is possible. Later, we will see concrete examples of this with commutators $[X,Y]$, which must be multiplied by $i$ to turn Hermitian. We will also see more fundamental reasons to choose complex numbers in QM.



Exercise: Show that the expected value of an observable $X$ given a state $\psi$ can be given as $\langle \psi|X|\psi \rangle$ (i.e. $\psi^*X\psi$ in conventional notation).

Exercise: Explain Born's rule with other observables, like position and momentum. Explain why it holds in general.

Invariant and generalised eigenspaces; Jordan normal form

We defined "eigenvectors" -- or really "eigenlines" -- in order to understand the behaviour of linear transformations as scalings across certain axes (which may be complex, and the scalings may be complex too). But simply thinking of eigenlines as 1-dimensional spaces that a transformation leaves invariant (the fancy phrase here is: "the 1-dimensional subspaces on which it is an endomorphism"), it is natural to wonder about higher-dimensional invariant spaces -- subspaces on which some transformation $A$ acts as an endomorphism.

The problem is that any transformation has all sorts of useless invariant subspaces -- for instance any transformation $A:F^n\to F^m$ (where $m \le n$) has the entirety of $F^n$ as an invariant subspace (for any $F$), and rotations -- although fully described by their eigenvectors and eigenvalues -- have a bunch of real and complex planes as unnecessary invariant subspaces. And if $A$ has an eigenvalue with geometric multiplicity $>1$, there are an infinite number of useless invariant subspaces.


Specifically, if the goal is to find useful representations of defective matrices (non-diagonalisable), invariant subspaces seem completely useless -- they certainly have no hope of giving us any sort of unique representation. Perhaps more on the point, our "eigenlines" have corresponding eigenvalues that tell us how the transformation behaves within an eigenline. Our invariant subspaces currently have nothing of the sort -- the transformation can have any sort of behaviour on the invariant subspace -- rotation, skewing/scaling, shearing, skewering -- and we'd have no idea. We need a convenient way to write down the behaviour of the transformation on an invariant subspace.

Here's something we can start to think about: ordinary eigenvectors satisfy $(A-\lambda I)v=0$, which gives us a one-dimensional solution space. In analogy with solutions to linear differential equations (linear homogenous if you use the conventional terminology, but I reserve "affine" for linear non-homogenous), an equation like

$$(A-\lambda_1 I)(A-\lambda_2 I)v=0$$
(where ${\lambda _1},{\lambda _2}$ are both eigenvalues of $A$) would have a 2-dimensional solution space, etc.

Note that when $\lambda_1, \lambda_2$ are not eigenvalues, we don't have a 2-dimensional solution space (what does the solution space look like then?). Why does it work with differential equations for any $\lambda_1, \lambda_2$? (hint: what do the eigenvalues look like?)

It's sensible to ask: are these solution spaces the same as our invariant subspaces? I.e. is every member of a $k$-dimensional invariant subspace a solution to an equation of the form

$$(A - {\lambda _1}I)...(A - {\lambda _k}I)v = 0$$
for eigenvalues $\lambda_1,...\lambda_k$?

The answer is yes. I encourage you to try and prove it for yourself -- it is instructive to first consider special cases: (i) rotation in a plane, where indeed $(A-iI)(A+iI)=A^2+1=0$ is the minimal polynomial of $A$ (ii) more generally, $F^n$ is an invariant subspace for all isomorphisms, and indeed for all $v$ in this subspace (i.e. $\forall v \in F^n$), $p(A)v=0$ where $p$ is the characteristic polynomial of $A$ by the Cayley-Hamilton theorem.

The key to proving that every invariant subspace is given by solutions to an equation of the form $(A - {\lambda _1}I)...(A - {\lambda _k}I)v = 0$ (and vice versa) lies in recognising that on any $k$-dimensional invariant subspace, $A$ is acts as an endomorphism, and therefore the Cayley-Hamilton theorem applies to it, with a $k$-order characteristic polynomial.

I encourage you to spend some time thinking about this -- try relating it to differential equations. Come up with another proof of the statement -- an inductive one. See if this results in a better intuition for the Cayley-Hamilton theorem.

Is it true that there $2^n$ invariant subspaces of any transformation on an $n$-dimensional linear space? What about the identity transformation?



Now, let's discard the invariant subspaces we don't want. We already know how to handle cases with distinct eigenvalues -- i.e. we have distinct eigenvalues in $\lambda_1...\lambda_k$ -- we just get an eigenvector for each eigenvalue. So we're really just concerned with subspaces of the form ${(A - \lambda I)^k}v = 0$. This is analogous to linear differential equations with repeated roots being weirder than ones with distinct roots.

Note that we still know how to handle ${(A - \lambda I)^k}v = 0$-like equations when the algebraic multiplicity is accounted for by geometric multiplicity -- when this is the case, you can reduce the power from $k$ (by subtracting from it the geometric multiplicity). This nuance doesn't exist with differential equations, because distinct eigenvectors have distinct eigenvalues.

Vectors satisfying such an equation are called generalised eigenvectors of order $k$ where $k$ is the minimum value for which it does satisfy the equation, and the invariant subspaces formed by generalised eigenvectors of the same eigenvalue are called generalised eigenspaces. The dimension of the generalised eigenspace always equals the algebraic multiplicity, unlike the eigenspace, whose dimension equals the geometric multiplicity.

Check this, and that $k$ is the difference between algebraic and geometric multiplicity.

What kind of transformations precisely do generalised eigenvectors with degree greater than 1 correspond to? Clearly, skews and rotations are out of the question. But some insight can be gained from looking at the nature of skews and rotations on a plane.


In two dimensions, a characteristic polynomial with a positive discriminant yields a skew along some axis, a negative discriminant yields a rotation, and the case we're interested -- the presence of repeated roots -- corresponds to the point "between" skews and rotations (speaking hand-wavily), shears.

In a more general setting, if one has: ${(A - \lambda I)^k}v_k = 0$  for generalised eigenvector  $v_k$of degree $k$, then one can extract generalised eigenvectors of each degree lower:

$$\begin{array}{l}{(A - \lambda )^i}{(A - \lambda )^{k - i}}{v_k} = 0
\\ \Rightarrow {v_i} = {(A - \lambda I)^{k - i}}{v_k}\end{array}$$
Implying that the generalised eigenvectors with the same eigenvalue (can) form a basis for the corresponding generalised eigenspace.

$$\begin{array}{*{20}{r}}{(A - \lambda I){v_1} = 0 \Leftrightarrow A{v_1} = \lambda {v_1}}\\{{{(A - \lambda I)}^2}{v_2} = 0 \Leftarrow (A - \lambda I){v_2} = {c_1}{v_1} \Leftrightarrow A{v_2} = {c_1}{v_1} + \lambda {v_2}}\\{{{(A - \lambda I)}^3}{v_3} = 0 \Leftarrow (A - \lambda I){v_3} = {c_2}{v_2} \Leftrightarrow A{v_3} = {c_2}{v_2} + \lambda {v_3}}\\{ \vdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{{(A - \lambda I)}^k}{v_k} = 0 \Leftarrow (A - \lambda I){v_k} = {c_{k - 1}}{v_{k - 1}} \Leftrightarrow A{v_k} = {c_{k - 1}}{v_{k - 1}} + \lambda {v_k}}\end{array}$$
This gives us a very clear picture of how these general "sheary" transformations look in the basis of the generalised eigenvectors -- there is a shear in each plane $\left\langle {{v_{k - 1}},{v_k}} \right\rangle$. Suitable scalings could of course be chosen to make all the $c_i=1$.

It's hard to overstate the significance of this -- what we've just found is that any defective matrix can be decomposed into shears on each of its generalised eigenspaces. This completely classifies the diagonalisability of matrices. If it's defective, it's a shear.

Draw out some of these transformations in three or more dimensions.

Notice the directions of the implication signs -- can we make them double-sided? What if we have some geometric multiplicity? How would our shears look like then? How many dimensions do you need to visualise this?

Think about why this characterisation of defective matrices makes sense. What effect does adding a 1 to the subdiagonal have on the determinant? Why? (hint: area of a parallelogram) What about the other coefficients of the characteristic polynomial? (hint: think of these in terms of traces of some matrix). So these matrices are precisely those which have the same characteristic polynomial as a diagonalisable matrix without actually being similar to one.



Clearly, the generalised eigenspaces of a transformation are pairwise disjoint (i.e. intersect only at the origin). Since all eigenvalues are being considered, their union is all of $F^n$. Thus the union of their bases forms a basis for $F^n$. This gives us a representation of the transformation $A$ in this basis.

From the last section, it is clear that the effective transformation on a $k$-dimensional eigenspace with eigenvalue $\lambda$ (called a "Jordan block") is given by:

$$\left[ {\begin{array}{*{20}{c}}\lambda &0&0& \cdots &0\\1&\lambda &0& \cdots &0\\0&1&\lambda & \cdots &0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\0&0&0& \cdots &\lambda \end{array}} \right]$$
The Jordan normal form of a transformation is then the matrix formed by putting all Jordan normal blocks along the diagonal -- i.e. the representation of $A$ in its generalised eigenbasis.

Some writers define the Jordan normal form with the 1's on the superdiagonal. It should be clear to you from the work we've done that this form is obtained by taking the basis vectors in reverse order (i.e. changing the basis to $\langle e_n,...,e_1 \rangle$. If this isn't clear to you, go back and work through the previous section once more. Loop.

(Stay tuned for the next article to see how we can -- instead of defining generalised eigenspaces whose dimension is defined by the algebraic, rather than geometric multiplicity -- "force" algebraic multiplicity to equal geometric multiplicity, i.e. diagonalise any matrix -- with a ring extension.)

All matrices can be diagonalised over R[X]/(X^n)

This post follows from my answer to the math stackexchange question What kind of matrices are non-diagonalisable?


Non-diagonalisable 2 by 2 matrices can be diagonalised over the dual numbers -- and the "weird cases" like the Galilean transformation are not fundamentally different from the nilpotent matrices.

The intuition here is that the Galilean transformation is sort of a "boundary case" between real-diagonalisability (skews) and complex-diagonalisability (rotations) (which you can sort of think in terms of discriminants). In the case of the Galilean transformation $\left[\begin{array}{*{20}{c}}{1}&{v}\\{0}&{1}\end{array}\right]$, it's a small perturbation away from being diagonalisable, i.e. it sort of has "repeated eigenvectors" (you can visualise this with MatVis). So one may imagine that the two eigenvectors are only an "epsilon" away, where $\varepsilon$ is the unit dual satisfying $\varepsilon^2=0$ (called the "soul"). Indeed, its characteristic polynomial is:

$$(\lambda-1)^2=0$$
Whose solutions among the dual numbers are $\lambda=1+k\varepsilon$ for real $k$. So one may "diagonalise" the Galilean transformation over the dual numbers as e.g.:

$$\left[\begin{array}{*{20}{c}}{1}&{0}\\{0}&{1+v\varepsilon}\end{array}\right]$$
Granted this is not unique, this is formed from the change-of-basis matrix $\left[\begin{array}{*{20}{c}}{1}&{1}\\{0}&{\epsilon}\end{array}\right]$, but any vector of the form $(1,k\varepsilon)$ is a valid eigenvector. You could, if you like, consider this a canonical or "principal value" of the diagonalisation, and in general each diagonalisation corresponds to a limit you can take of real/complex-diagonalisable transformations. Another way of thinking about this is that there is an entire eigenspace spanned by $(1,0)$ and $(1,\varepsilon)$ in that little gap of multiplicity. In this sense, the geometric multiplicity is forced to be equal to the algebraic multiplicity*.

Then a nilpotent matrix with characteristic polynomial $\lambda^2=0$ has solutions $\lambda=k\varepsilon$, and is simply diagonalised as:

$$\left[\begin{array}{*{20}{c}}{0}&{0}\\{0}&{\varepsilon}\end{array}\right]$$
(Think about this.) Indeed, the resulting matrix has minimal polynomial $\lambda^2=0$, and the eigenvectors are as before.



What about higher dimensional matrices? Consider:

$$\left[ {\begin{array}{*{20}{c}}0&v&0\\0&0&w\\0&0&0\end{array}} \right]$$
This is a nilpotent matrix $A$ satisfying $A^3=0$ (but not $A^2=0$). The characteristic polynomial is $\lambda^3=0$. Although $\varepsilon$ might seem like a sensible choice, it doesn't really do the trick -- if you try a diagonalisation of the form $\mathrm{diag}(0,v\varepsilon,w\varepsilon)$, it has minimal polynomial $A^2=0$, which is wrong. Indeed, you won't be able to find three linearly independent eigenvectors to diagonalise the matrix this way -- they'll all take the form $(a+b\varepsilon,0,0)$.

Instead, you need to consider a generalisation of the dual numbers, sometimes called (in computing mathematics and non-standard analysis) the "hyperdual numbers", with the soul satisfying $\epsilon^n=0$. Then the diagonalisation takes for instance the form:

$$\left[ {\begin{array}{*{20}{c}}0&0&0\\0&{v\epsilon}&0\\0&0&{w\epsilon}\end{array}} \right]$$


*Over the reals and complexes, when one defines algebraic multiplicity (as "the multiplicity of the corresponding factor in the characteristic polynomial"), there is a single eigenvalue corresponding to that factor. This is of course no longer true over the hyperdual numbers, because they are not a field, and $ab=0$ no longer implies "$a=0$ or $b=0$".

In general, if you want to prove things about these numbers, the way to formalise them is by constructing them as the quotient $\mathbb{R}[X]/(X^n)$, so you actually have something clear to work with.

(Perhaps relevant: Grassmann numbers as eigenvalues of nilpotent operators -- the Hyperdual numbers are not the same as the Grassmann numbers, and the algebra of the Grassmann numbers is definitely different from that of nilpotent and shear matrices, but go see if you can make sense of it.)

Something important to note is that the diagonalisation is not of the form $D=P^{-1}AP$, as the eigenvector matrices are not invertible. However, it is still true that $PD=AP$ -- nonetheless, this limitation prevents this formalism for being any good for e.g. dealing with polynomial-ish differential equations with repeated roots, for instance, as far as I can see. The infinitesimal-perturbation/"take a limit" approach we talked about in Limiting Cases II: repeated roots of a differential equation are still the right approach for that.

Null, row spaces, transpose, fundamental theorem of linear algebra

This is an exciting article, so pay close attention.

In the last article, we introduced the column space, and we are lead to wonder about the vectors that are mapped to the zero vector (let's call the set of these vectors the "null space"). Based on some of the transformations we've seen, we might wonder if the null space is essentially the subspace perpendicular to the column space, i.e. the set of all vectors perpendicular to every vector in the column space (we call this the "orthogonal complement" of the column space).

However, this is demonstrably wrong. Suppose one rotates the co-ordinate system  before collapsing it onto a lower dimension. The "collapsing" matrix, the singular matrix, may be $\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]$, and the rotation matrix may be $\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right]$, then the composite transformation is $\left[ {\begin{array}{*{20}{c}}1&0\\0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0&{ - 1}\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0&{ - 1}\\0&0\end{array}} \right]$

In general in $\mathbb{R}^2$, if the "simple collapse" transformation looks like this:

$$\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\cos \phi \sin \phi }\\{\cos \phi \sin \phi }&{{{\sin }^2}\phi }\end{array}} \right]$$
(Show that the non-rotating collapses can be written in this way), and the rotation matrix looks like this:

$$\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]$$
Then their product

$$\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\phi }&{\cos \phi \sin \phi }\\{\cos \phi \sin \phi }&{{{\sin }^2}\phi }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right] \\= \left[ {\begin{array}{*{20}{c}}{\cos \phi \cos (\phi  - \theta )}&{\cos \phi \sin (\phi  - \theta )}\\{\sin \phi \cos (\phi  - \theta )}&{\sin \phi \sin (\phi  - \theta )}\end{array}} \right]$$
Which is also a singular matrix with the same rank and column space as the original, but its null space is a rotated version of the original, and therefore no longer perpendicular to the column space.

We make the following observations:
  1. The rotation has introduced an "asymmetry" in the components of the matrix -- the matrix is no longer symmetric across the principal diagonal, the rows and columns are no longer the same. In our original terminology before we introduced matrices, the "contribution of y to x" is no longer the same as "the contribution of x to y. 
  2. The span of the column vectors is not dependent on $\theta$ (proving again that the column space is not the space orthogonal to the null space), but the span of the row vectors is -- we'll call this the row space. The row space used to be the same as the column space, but is now rotated clockwise by an angle of $\theta$. Since the space is rotated counter-clockwise by $\theta$ before being collapsed, this means the null space is $\theta$ clockwise to what its position would have been without the rotation. 
Hence, the null space is perpendicular to the row space, at least if the pre-collapse transformation is a rotation. It would be a very educated guess to say that this applies generally, that the null space of a transformation is always perpendicular to the row space.

Indeed, this is the case, and is pretty easy to show:

$$\left[ {\begin{array}{*{20}{c}}{{r_1}}\\ \vdots \\{{r_n}}\end{array}} \right]\vec x = 0\,\,\, \Rightarrow \,\,\,{r_i} \cdot \vec x = 0\,\,\,\,\,\,\forall i \in [1,n] \cap \mathbb{Z}$$
More interestingly, however, we made a pretty important observation about the symmetry of a matrix: asymmetry in the matrix seems to be a measure of "rotation-ish" a matrix is. The reason this makes sense is that while the values on the principal diagonal talk about how much each component is scaled in the resulting image, the off-diagonal elements talk about how much contribution there is to one component of a vector from the component in another direction, something that happens during rotations.

If the contribution from x to y is exactly the same as the contribution from y to x, then the xy-plane isn't really being rotated, but the basis vectors are rather being pulled closer to/apart from each other. On the other hand, when they are different, the "orientation" of the axes changes, and an actual rotation is induced.

The most "rotation-ish" effect is created when the values of the matrix are the negative of the their reflection across the principal diagonal, because it means the basis vectors are being rotated by the same angle in the same direction.

Quick exercise: plot how such a transformation looks in $\mathbb{R}^2$. You'll notice that the rotation isn't as "perfect" as you might've hoped -- the vectors change length, etc., and it's the orthogonal displacement in the basis vectors that is the same in orthogonal directions for the basis vectors, not the angles.

These changes in length are a result of scaling, i.e. the values along the principal diagonal.

Well, not exactly, because the values along the principal diagonal scale the part of the basis vectors in their original direction, not the overall length of the basis vectors. This is why (i) the resulting matrix not only eliminates the scaling, but also ensures the rotation is in right-angles, (ii) the resulting matrix is not actually just a bunch of pure rotation, as its rotations are still scaled, and (iii) why some otherwise-antisymmetric matrices with a principal diagonal, like$\left[ {\begin{array}{*{20}{c}}
  {\cos \theta }&{ - \sin \theta } \\
  {\sin \theta }&{\cos \theta }
\end{array}} \right]$, can still be rotations in some angle other than a right angle. Highly recommended reading: my answer to "Intuition behind the Speciality of Symmetric Matrices".

Therefore, it is useful to extract from the matrix the purely anti-symmetric part, with a zero principal diagonal:

$$\left[ {\begin{array}{*{20}{c}}0&{{b_{12}}}& \ldots &{{b_{1n}}}\\{ - {b_{12}}}&0& \ldots &{{b_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{ - {b_{1n}}}&{ - {b_{2n}}}& \ldots &0\end{array}} \right]$$
This matrix is called a skew-symmetric matrix or an anti-symmetric matrix, and in $\mathbb{R}^n$ is essentially a combinations of scaled rotations around different axes.

Meanwhile, one may define a symmetric matrix in the following way:

$$\left[ {\begin{array}{*{20}{c}}b_{11}&{{b_{12}}}& \ldots &{{b_{1n}}}\\{{b_{12}}}&b_{12}& \ldots &{{b_{2n}}}\\ \vdots & \vdots & \ddots & \vdots \\{{b_{1n}}}&{{b_{2n}}}& \ldots &b_{nn}\end{array}} \right]$$
These matrices essentially scale and skew vectors -- the principal diagonal components do the scaling in each direction, and the other components do the skewing, i.e. tilting the basis vectors towards each other.

One may wonder if it would have been better to simply deal with diagonal matrices instead of symmetric ones. However, as we will see, it turns out that any matrix can be written as the sum of a symmetric matrix and an anti-symmetric matrix. It is also worth noting that we will eventually see that a symmetric matrix is a generalisation of a scaling matrix, where the scaling is done in some arbitrary directions that need not be the basis vectors. These vectors are called "eigenvectors".

Looking at the definitions of symmetric and antisymmetric matrices, it is clear that any matrix may be written as the sum of a symmetric matrix and an antisymmetric one. Specifically, one may write:

$$A = \underbrace {\frac{1}{2}(A + {A^T})}_{\scriptstyle{\rm{symmetric }}\atop\scriptstyle{\rm{part}}} + \underbrace {\frac{1}{2}(A - {A^T})}_{\scriptstyle{\rm{antisymmetric }}\atop\scriptstyle{\rm{part}}}$$
Where $A^T$ is the "transpose" of $A$, referring to the matrix formed upon taking flipping $A$'s entries across the principal diagonal. E.g. the row space of $A$ is the column space of $A^T$.

Does this all remind you of something?
  • Any entity can be written in two "parts" of a specific nature, which look curiously like the exponential form of the cosine and sine functions
  • These two parts represent scaling and scaled $\pi/2$ rotation respectively.
It should remind you of the Cartesian form of a complex number.

A symmetric matrix is "like" a real number, and an anti-symmetric matrix is "like" an imaginary number. A complex number can be thought of as an object to be transformed (like a vector) as well as as a transformation itself (like a matrix).


However, it is important to highlight the differences: (i) while a real number only scales things in their own direction (like a multiple of $I$), a symmetric matrix can skew things, which is equivalent to scaling across a different set of axes. Therefore while complex numbers can encode spiral transformations, matrices encode all linear transformations. (ii) complex numbers transform on the complex plane, which is a two dimensional plane like $\mathbb{R}^2$, while linear transformations can operate in any number of dimensions, and not necessarily even just in $\mathbb{R}^n$. An example of a consequence of this would be that an anti-symmetric matrix can encode a series of right-angle rotations with corresponding scalings in dimensions greater than three, but an imaginary number can only correspond to a rotation around an axis pointing out of the plane.

It is fair to say that linear algebra generalises complex numbers with matrices. For some examples of correspondence between specific complex numbers, see Introduction to symmetry, section "symmetry on the complex plane". We will formalise this whole idea of matrices being "like real numbers" and "like imaginary numbers" when we do eigenvalues and eigenbases -- in fact, antisymmetric matrices have imaginary eigenvalues.

The complex conjugate, similarly, is generalised to the transpose. Explain how. (Hint: $A^T$ does precisely the opposite rotation-ish, i.e. antisymmetric actions as $A$ while preserving the symmetric actions. What does this tell you about matrices whose transpose and inverse are equal? What does it tell you about matrices that are equal to their transpose?)


Watch the above Khan Academy video.
Come up with an intuitive explanation for why the row space solution is the shortest.


The three following results:
  1. Row space is perpendicular to the null space
  2. Row rank equals column rank
  3. Rank plus nullity equals dimension (Rank-nullity theorem)
Together imply what is called the fundamental theorem of linear algebra -- that the row space is the orthogonal complement of the null space. Why? To be the orthogonal complement, you need to be orthogonal (implied by 1), and your dimensions need to add up to the total dimension of the space. The latter is implied by 3, but our intuition applies perhaps better to the dimension of the column space, and 2 implies that this has the same dimension as the row space, and we're done.

Among these three results, this article provides fairly solid intuition for the first and the third. The second is really assumed throughout this article, but if you want really solid intuition for it, read on.

Consider an $m$ by $n$ matrix $A$ with linearly independent rows (i.e. its row rank is $m$) (we can always turn matrices to this form with a bunch of zero rows at the end via row operations). Then the nullity of $A$, i.e. the number of solutions to $Ax=0$ is $n-m$, as there are $m$ equations in $n$ variables. Thus we have that row rank plus nullity equals the dimension. But we know from our "collapsing intuition" that column rank plus nullity equals the dimension. And we're done.

An obvious consequence of rank-nullity is that injectivity is equivalent to surjectivity. However, this is not true for infinite-dimensional spaces. Why not? You might find it's related to Hilbert's hotel.



It's worth noting that while we've pretended that having the column space equal to the row space is equivalent to being symmetric, this is not really true. There's a broader set of matrices for which the column and row spaces are equal, called "rank-symmetric" or EP matrices.

This article is extended in SVD, polar decomposition, normal matrices; a re-look at transposes and FTLA.