(Snapping out of it) Okay, so uh.....advancing further along the algebra path, we then start to see square roots of not so "perfect" numbers (e.g. 8, 14, 39, 45). Unlike 4 or 9 we can't find whole numbers that when squared will match these exactly. It won't work! Never ones to admit defeat, we roll up our sleeves, manipulate 'em radicals, and figure out how to at least simplify said square roots (or use our calculators if all else fails! not recommended though :-P.....I MEAN IT). That's fine and most of us will then choose to stop our little "land of square roots" family vacation there. There are always a few brave souls though with a taste for danger who will decide to venture on (show-offs). First stop: the "imaginary sulfur pits" (ominous and foreboding music starting to play). Below is an account recorded by one of these adventurers. His name is not mentioned as he would like to protect his anonymity. :-) Recording:
Well there's nothing even trying to be imaginary about the smell. I'm catching a strong whiff of something resembling eggs (makes me think of my Aunt Edna's egg salad...uggh). The smoke's starting to depart now. There appear to be several piles of rocks surrounding the yellowish pit, each inscribed with various symbols: a bunch of lower case i's and -1's, the latter all surrounded by square root symbols. The writing looks pretty old. Were these ruins created by a long lost civilization? If so what was with their fixation on the 9th letter of the alphabet? Even more disturbing though: did the ancients know how to do their math because they keep on trying to take the square root of a negative number: impossible! Hmmmmm....what to do? I think I'll pull out my trusty math guidebook from the glove department. It probably has something to say about this.
(flipping pages)
Ah! Here we are! Imaginary numbers (explains the "imaginary" sulfur pits at least). And look here! It says that you can define a whole new set of numbers as an extension to the real numbers, denoting them collectively as C. Within this numbering system it is possible to take the square roots of negative numbers, the very base case being the square root of -1. Rather than writing -1 with a square root symbol over and over again it recommends denoting the square root of -1 as i. Helps to make simplifying complex radicals easier I suppose. Oh and look hun! There's even a cute little example here where it shows you the square root of -4: 2i and -2i. I get there being two square roots but this still seems a little weird. Hughh.... there's even a coordinate system on the next page where the x and y axes are also complex. How odd! It says for further information either directly ask for a park ranger who knows complex analysis or go read a freakin' textbook! How rude!
This family vacation keeps on getting curiouser and curiouser (Alice in Wonderland moment). The sulfur pits were alright, a little different for me to get my mind around. Kind of cool though. What do you say kids? You ready to call it a day or did you still want to check out the museum of petrified matrix roots? (Loud protesting noises and shouts of excitement in the background). Ahhh very well. Onto the museum then.
(fast forwarding tape recorder)
And here we are! Final destination stop: the museum. It looks nice enough. There are display cases everywhere filled with samples of petrified matrix roots. In the center of each display case is a black and white picture of a diagonalizable matrix. Below said pictures are roots that have been supposedly extracted from the matrices.The museum lady says that they're all from the forest of Diag'ble matrices, donated by generous mathematician patrons from around the world. Let's see how many there are: 1, 2, 3, 4, oh my goodness! 5, 6, 7, and 8!!! I thought that matrices just had two roots, you know, like the stuff at the sulfur pits and the radicals shown on the signs leading up to there. How on earth could something have more than 2 roots? Hmmmnnnn.... the sign by this display case might give a hint; or rather a whole explanation. It reads as follows:
Square Roots of Matrices:
The square root of a square matrix A is a matrix B such that when you square B you get A. How does one find these square roots and how many of them are there? The first temptation that many might have is to simply take the square root of each of the entries within matrix A and call it a day. If you test out your answer though you will see that in general due to the way matrix multiplication is defined, this approach will not give you back your original matrix A. A way to get to the square root of the matrices shown here is a bit more involved than that.
The matrix shown in the above picture was taken from the forest of Diag'ble matrices in the year 1895. Although it is quite old, due to the preservation techniques of volunteering mathematicians it has maintained all of its properties since then, namely that it is still diagonalizable. This means that it can be written as the product of three matrices: (S^-1)*D*S (or equivalently as the product T*D*(T^-1)for some matrix T and diagonal matrix D). Here S is a matrix that has an inverse (denoted as S^-1) and D is a diagonal matrix that just has zero entries off of the main diagonal (diagonal entries may be zero as well).
It is keeping this form in mind that one can then extract roots like the petrified ones displayed below said picture. To find them you take the following steps:
1. Find the eigenvalues of matrix A.
2. Find eigenvectors that correspond to the eigenvalues of #1.
3. Form S from the eigenvectors of #2.These eignenvectos will be the columns of S.
4. Compute the inverse of S.
5. The diagonal entries of D will correspond to the eigenvalues of A. Their ordering on the main diagonal will in turn depend on the ordering of the eigenvectors in S. E.g. if eigenvector x1 is put into column 1 of S, then its associated eigenvalue will be the first main diagonal entry of D. If eigenvector x2 is designated as the second column of S, then its eigenvalue will be the second diagonal entry in D,.., and so on.
6. Find a square root of D and denote it as D*. Calculating a root is very simple in this case: because of the form of diagonal matrices you can simply take square roots of all of D's entries. Since each nonzero number within D can have two real square roots, this can potentially create up to n^2 different real D*'s.
7. SD*(S^-1) will all qualify as square roots of matrix A. Notice that there will be as many roots of this form as there are roots of D, explaining the case of having at least 8 square roots.
End of plaque explanation
Well that makes some more sense when explained like that. Pretty amazing that you can have more than 2 square roots for a matrix! And I thought that all of that facebook and twitter stuff was crazy! What do you think about all of this kids?.....Kids?
Ahhh...looks like they've gone straight to the gift shops. Better go chase 'em down before they try and destroy everything.
End of Recording
Well that was errghh..enlightening. I decided to leave the complex roots part to the park rangers and go on with providing a bit more explanation of the matrix roots discussion through example. Here is a 3 by 3 matrix that I grew for myself out of a Diag'ble tree seedling I found myself. :-) Following the steps outlined above let's see if we can spot some of its roots!
First of all here is the lovely 3 by 3 matrix:
A = [9 -5 3]
[0 4 3]
[0 0 1]
Since the matrix is triangular its eigenvalues are the numbers along its main diagonal: 9, 4, and 1.
With a little bit of calculation involving the equation (cI - A)x = 0 (c denotes one of the above 3 eigenvalues, x is an eigenvector), you will find that the following 3 eigenvectors work (I will denote them as x1, x2 and x3).
x1 = [1 0 0]
x2 = [1 1 0]
x3 = [-1 -1 1]
For the matrix S I'll let the columns from left to right be the eigenvectors x1, x2 and x3 correspondingly.
S = [1 1 -1]
[0 1 1]
[0 0 1]
I did enough paper and pencil matrix calculations this past week, so I'll just let you know (and trust me it's correct) that the inverse of this matrix S is the following (to find the inverse you can form the augmented matrix from S and the identity and successively do row operations on both sides until the left side is the identity matrix).
S^-1 = [1 -1 2]
[0 1 -1]
[0 0 1]
We're almost done now! The diagonal matrix D will have the eigenvalues 9, 4 and 1 placed in order according to the order of the eigenvectors within S. As mentioned earlier a square root of D can be obtained by taking the square roots of each of its diagonal entries.
D = [9 0 0]
[0 4 0]
[0 0 2]
Example of a square root of D:
D* = [3 0 0]
[0 2 0]
[0 0 1]
Notice that there is not one unique square root of our diagonal matrix D; each of the elements along the main diagonal of our sample square root can be positive or negative. When squared they will all yield the same result D. Thus there are at least 8 roots of matrix D which in turn implies that there are at least 8 roots for A (again this is because when you square S*(D*)*(S^-1) you get back S*D*(S^-1), A diagonalizing representation of A). Pretty cool! :-D Now is anyone up for checking out those imaginary sulfur pits? ;D
