Problem: Find the standard matrix for the linear transformation which reflects points in the x-y plane across the line .

Solution: To find the matrix representing a given linear transformation all we need to do is to figure out where the basis vectors, i.e.

and

are mapped under the transformation.

This is easy to see from the fact that since and form a basis of any element

can be written as

for some .

Now let

be a linear transformation. Then, by definition

Thus, the image of any element under a linear transformation is completely determined by the image of the basis of the transformation’s domain.

Let’s find the image of under a reflection across the line given by . First, we need to find a line perpendicular to that passed through the point .

In general, a line perpendicular to is given by , i.e in our case, . To find the one particular line with the same slope, that passes through , we need to simply plug in the values of and into the line equation and add a constant value so that the equality will hold. Then we have

Thus, the line perpendicular to that passes through is given by

The lines given by equations (1) - red and (2) - green

Now let’s find the point at which our two lines and intersect. This is done easily by solving the two line equation simultaneously:

Equating and (and multiplying by 6 for simplicity) gives

which then gives

Thus, our two lines intersect at

Now, let be the reflection of across .  Since and are symmetric across (because that is how the reflection is actually constructed) it follows that

so

Similarly as above, we find the image of under a reflection across the line given by . The line perpendicular to that passed through the point is given by

The lines given by equations (1) - red, (2) - green, and (3) - blue

Equating and (and multiplying by 6 again) gives

and so

Therefore, and intersect at

and the image of under the reflection across is given by

which yields

Finally, the matrix for the linear transformation which reflects points in the x-y plane across the line is given by

It turns out (although I’m not going to show to derivation here) that in general, the matrix for a linear transformation which reflects points in the x-y plane across an arbitrary line is given by

where is the angle that the line makes with the positive x-axis, i.e

if

and

if

]]> http://www.scibuff.com/2009/06/22/reflection-matrix/feed/ 3 http://www.scibuff.com/2009/06/03/common-denominators/ http://www.scibuff.com/2009/06/03/common-denominators/#comments Wed, 03 Jun 2009 13:05:31 +0000 scibuff http://www.scibuff.com/?p=437 In the period of finals ( and studying for finals ) yahoo answers have been flooded with homework and assignment problems 99.9% involve mundane “plug it into a formula” type of solutions. As such, these are of no particular interest to me or for this series. Hopefully, now that most students swapped book for beaches, there will be more interesting questions posted. Meanwhile, here are two short problems I picked some time ago

Problem: Prove that if is relatively prime to and ,then

Solution: Using it follows

Since is relatively prime to we know that

and it must be true that

Thus .

Problem: If and prove .

Solution: Since we know that for some . Similarly, for some . Notice, that both and must be odd. If, lets say, were even, then for some . But then and so .

Using and we have . But since both and are odd, their sum is even, i.e.

for some

It now follows that

and so

Problem: Show set of real numbers between 0 and 1 with decimal representations consisting of 1s or 9s, i.e. is uncountable.

Solution: To solve this problem we will use a proof by contradiction just as we did before.

Suppose, that similarly as in the last problem, we can put our into a one-to-one correspondence with the set of natural numbers . In other words, we can create a list of ALL elements in our set (just as we did before) like so:

Now, consider any such arbitrary listing of the elements of , for example, this one:

For the following argument the order of the elements on the list is irrelevant. The important thing is we have listed them ALL. Now, suppose with create a number in the following way:

• Make the first (decimal) digit of be different from the first (decimal) digit of , i.e. 9
• Then make the second (decimal) digit of be different from the second (decimal) digit of , 1
• The third (decimal) digit of to be different from the third (decimal) digit of , i.e. 9
• The fourth(decimal) digit of to be different from the fourth (decimal) digit of , i.e. 1
• etc …

It is relatively easy to show that a number created in such a way as CANNOT be on our list. If were on the list then for some . So, in particular, the n-th (decimal) digit of must be equal to the n-th (decimal) digit of . But that is not possible by the very definition of .

On the other hand, set is the set of all real numbers between 0 and 1 with decimal representations consisting of 1s or 9s, in particular, .

The assumption that is countable led us to conclude that and at the same time , which is impossible. Therefore must be uncountable.

On a side note: can actually be put into one-to-one correspondence with the entire set of real numbers (which also true of any open interval of non-zero length)

Although he was definitely incorrect in his thinking and reasoning, because division be zero has no meaning when it comes to real numbers (or integers), there are cases, such as  The Riemann Sphere or a one-element field where the multiplicative identity coincides with the additive identity, when the equation above is actually true (and well defined).

It is sometimes quite tempting (and even useful) to consider

The main problem with allowing the division by zero is that it results in logical fallacies such as 1=0 (actually, one can equate any two numbers using it). Eventually, mathematicians found a way to define limits rigorously without infinitesimal quantities needed by Newton and Leibnitz when Cauchy and Weierstrass laid down the foundation of modern analysis.

The next question from Yahoo Answers is related to a different kind of infinity – the kind that deals with sets. The works of Georg Cantor between 1874 and 1884 are the origin of set theory in which he established the importance of one-to-one correspondence between sets, defined infinite and well-ordered sets, and proved that the real numbers are “more infinite” than the natural numbers. The diagonalization argument he used demonstrates a powerful and general technique that has since been used in a wide range of proofs such as this one.

Problem: Determine whether the set of real numbers between 0 and 1 with decimal representations consisting of 1s, i.e. is countable or uncountable?

Definitions:

• A function is bijective if and only if for every there is exactly one such that
• A set is said to be countable if there exists a bijective function where is the set of natural numbers.

Solution:

Let us look at set and consider the following list:

It is easy to see that our set can be easily put into a one-to-one correspondence with the set of natural numbers , e.g. the correspondence is given by the list (function defined) above.

To be precise, let be given by

To show that is bijective, we need to show that  for every there is exactly one, i.e. a unique, such that .

Consider an arbitrary . Since

we have

and so by taking we get

Therefore, for every we have proven the existance of (at least one) such that .

To show uniqueness, we need to demonstrate that if then . The easiest way to accomplish this is to use proof by contrapositive, that is, the fact that

is equivalent to

Let be such that . Without loss of generality, assume (otherwise rename them). Then it follows that there exist such that . Therefore

Since it follows that and so we have

thus proving the uniqueness. Therefore is a bijective function and so is countable.

Problem: Let be elements of group such that where is the identity element of . Prove that

Solution: The problem is indeed trivial if is abelian, i.e. for all . One must be careful not to assume such properties when dealing with constructs in abstract algebra. Nevertheless, the solution below is quite simple and uses only associativity of Groups, i.e. for all .

Lemma: For any there is a unique such that .

Proof: The lemma can be proven in two steps. First, show that such exists, and second, show that it must be unique.

Let . Since and is a group, it follows from the existence of  inverses that and from group closure . Thus, (and the first step is done).

Let be such that , take to be the identity element of . Then

Thus it follows that if then it must be that , finishing the second step, hence proving out lemma.

Now we go back to our proof. Suppose

is the identity in , i.e.

Using group associativity we get

“Multiplying” by (on right) we get

Now multiplying by (on left) we get

Now from we have

and from

Combining and

Since is the identity element of , by definition

Finally, using our Lemma and with it follows that

The beauty of the Fibonacci sequence and the golden ratio (which is intimately connected to it) lies in that they are not just another mathematical construct, but occur throughout the nature.

Have you ever taken a look at a pine cone and noticed that the scales of the cone are in spirals? Have you ever counted the spiral in the clockwise and counterclockwise directions? I would be surprised if you have … but the counts turn out to be 5 and 8 (or 8 and 13 for bigger cones).

Pine Cone Spirals

How about these? Care to count the petals?

13 petals

21 petals

… intriguing stuff indeed.

So, once again, I was reading through problems on Yahoo Answers. Most of the questions “reduce” (lol) to plugging numbers into well known equations, or using a calculator such as TI-89, or Mathematics, Maple, or even Google – basically, what engineers do. Yeah, you “heard” me right! It takes some effort to found a meaningful question that actually requires some knowledge and skills – see the difference between a mathematician and an engineer now? That said, here’s a one I couldn’t resist:

Problem: Find the term in the following sequence

There is a more “elegant” way to tackle this problem and I may write about in the future (maybe quite soon actually), but for now I’ll ignore matrices, diagonalization and eigenspaces (although the reason why the following solution gives the correct result is tightly connected to linear algebra) and focus, instead, on the recursive nature of the Fibonacci sequence.

Solution: To find the term, “all” that is needed is to find the closed form for the n-th terms of the sequence . In general, a Fibonacci sequence is given by

and any particular sequence is fully determined by the initial two terms

and

In our example, we have

and

To find the closed form, let’s assume

Then using our recursive formula we get

and so

Solving the quadratic from gives

Using this result, we can write the closed form formula for the n-th term of our sequence as

where and are constant parameters determined by our initial conditions,

and

We can calculate these values by solving the following system of equations

Plugging in known values gives us

Simplifying turns the above into

Now we need to solve and for and . Although I said I was going to leave matrices alone, this is a perfect situation to use them. I prefer to use the procedure described below for a system of two equations with coefficients such as these. Substitution or elimination method would, of course, work, but they involves messy arithmetic which can be so easily error prone in situations such as this one.

We have the following system:

In general, to solve the equation

we use

which turns into

Hence, to solve we need to find the inverse of our matrix. This is fairly simple for a 2 by 2 matrix … Let

then

and so

Our equation then becomes

which gives us

Finally, by plugging and into we get

We can now “easily” compute which is given by

Google returns which is just an approximation, but now we have a formula for the precise answer.

If you wish to check for yourself that out formula is indeed correct, click on the terms below to see the computation by Google:

On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me?

Consider this:

To calculate

we use the fact that

…

As you can see, in every step (multiplication) we use the result from the previous one and so it takes only 8 steps (multiplications) to calculate

Finally, in the process of calculating

we get

and

as a bonus, and so to calculate

we now only need to use two more multiplications

This idea can be generalized to any

The (maximal) number of multiplications required to compute

is given by

To put this result in a little bit of perspective, consider

This number is the currently largest known (Mersenne) prime. It has 12,978,189 digits! But and so

It follows that we can calculate

using at most 51 multiplications! Incredible!

Problem: Prove that there are no integer solutions to

Like so many problems in mathematics, this one is quite easy to solve if you know which tools to throw at it. Otherwise, one can always look at first few n’s, i.e. 1, 2, 3, 4, etc. and see if there is a pattern or anything that could be used in the proof.

When it comes to non-existence proofs involving perfect squares the obvious choice of “tools” is modular arithmetic. In this particular proof, I will also use proof by contradiction.

Proof by contradiction is a form of argument in which the original result is proven by disproving the opposite (formally known as the “negation” of the original statement). Basically, I will assume that there are two integers such that the equality above holds and use my “tools” to arrive at an obvious contradiction, such as .

Proof: First, let’s make our lives a bit easier by renaming the terms. If there are integers satisfying the equality in , then taking we can rename the terms to

Let’s assume that there are such that holds. Expanding the terms in brackets gives

Now look at and apply arithmetic modulo 3. Clearly

,

and so

.

Since

it follows that

This proof could be accomplished using modular arithmetic modulo 8 or 16 but the nice thing about modulo 3 is that there are only three possible remainders, namely 0, 1 and 2, in modulo 3 arithmetics (as opposed to 8 in modulo 8 and 16 in modulo 16 – you see the pattern there, right?!). Each remainder is associated with a congruence class – the set of all numbers that have the same remainder modulo 3, e.g. {1, 4, 7, 10, etc.}

We know that any integer belongs to one of the three congruence classes in modular arithmetic modulo 3. Now let us examine squares of integers and see to which congruence classes can those be assigned.

Consider such that . Then

Similarly, if , then

and finally, if , then

Thus, we can conclude that squares of all integers belong to either the congruence class or modulo 3. In particular, there is no integer whose square belongs to the congruence class modulo 3. In other words,

there is no such that

But that is exactly the opposite result of what we concluded in . Clearly, is correct, therefore must be false. We arrived at the result in by assuming that there are integers such that

Therefore, this assumption must be erronous, and the opposite must be true. Hence there are no integers such that

, or equivalently

Thus out proof is complete and we can finish it off in style by writing from latin Quod Erat Demonstrandum, which literally means “which was to be demonstrated”.