03/16/19

# What does the given fact that a matrix A times itself equals -A have to say about its eigenvalues?

My answer to a question on Quora: What does the given fact that a matrix A times itself equals -A have to say about its eigenvalues?

There were two answers in this thread, and both correctly stated that the eigenvalues of $$A$$ equal $$0$$ or $$-1$$. However, one of the answers contains the assertion:

They are all either zero or $$-1$$, regardless of $$A$$’s being diagonalizable or invertible.

I wish to add that $$A$$ is actually diagonalisable. And here is a proof.

Let $$A$$ be a $$n\times n$$ matrix. To prove that $$A$$ is diagonalisable, it suffices to show that the vector space $$V= \mathbb{R}^n$$ can be written as a sum of eigenspaces $$V_0$$ and $$V_{-1}$$ for $$A$$ and its eigenvalues $$0$$ and $$-1$$, correspondingly, or, which is the same, every vector $$\vec{v} \in V$$ can be presented as a sum

$$\vec{v} = \vec{v}_0 + \vec{v}_{-1}$$

with $$\vec{v}_0\in V_0$$ and $$\vec{v}_{-1}\in V_{-1}$$. But this is easy; indeed, obviously

$$\vec{v} = (\vec{v} + A\vec{v}) – A\vec{v}.$$

Observe that

$$A(\vec{v} + A\vec{v}) = A\vec{v} +A^2\vec{v} = A\vec{v} – A\vec{v} = 0$$

and

$$A(-A\vec{v}) = -A^2\vec{v} = -(-A\vec{v}),$$

so we can take

$$\vec{v}_0 = \vec{v} + A\vec{v} \quad \mbox{ and }\quad \vec{v}_{-1} = -A \vec{v}.$$

03/16/19

# Why do some square matrices not have an inverse?

My answer to a question on Quora: Why do some square matrices not have an inverse? I need a simple answer.

A good question, and it can be explained by a simple observations: there are non-zero square matrices $$A$$ and $$B$$ such that the product is the zero matrix: $$AB = 0$$. Let us look at the example given by Alexander Farrugia:

$$A = \left(\begin{array}{rr} 1&2\\ 1&2\end{array}\right)$$

it is not invertible because if I take

$$B = \left(\begin{array}{rr} -2&-2\\ 1&1\end{array}\right)$$

then their product is the zero matrix:

$$AB = \left(\begin{array}{rr} 1&2\\ 1&2\end{array}\right) \left(\begin{array}{rr} -2&-2\\ 1&1\end{array}\right)=\left(\begin{array}{rr} 0&0\\ 0&0\end{array}\right).$$

Can after that $$A$$ be inversible? No, it cannot, because multiplying the last equality by $$A^{-1}$$, we get $$A^{-1}AB = A^{-1}\cdot 0$$, which simplifies as $$B=0$$.

I will give this as a simple exercise to my (first-year) students:

A square matrix $$A$$ is either invertible, or there exists a non-zero square matrix $$B$$ such that $$AB = 0$$.

03/15/19

# Is there a reason matrix multiplication is defined as row times column and not row times row?

My answer to a question on Quora: Is there a reason matrix multiplication is defined as row times column and not row times row?

Let us look at a basic (and real life) example of matrix multiplication: a matrix made of a single row is multiplied by a matrix made of a single column. This example is calculation of the cost of purchase of some amounts $$g_1,g_2, g_3$$ of some goods (say, apples, bananas, and oranges) at prices $$p_1,p_2, p_3$$ of pounds (of sterling) per kilogram. The answer is well known:

$$\displaystyle{p_1g_1 + p_2 g_2 +p_3g_3 = \sum_{i=1}^3 p_ig_i},$$

which could be conveniently written as a matrix product:

$$\displaystyle{p_1g_1 + p_2 g_2 +p_3g_3 = \left[\begin{array}{ccc} p_1 & p_2 & p_3\end{array}\right]\cdot \left[\begin{array}{c} g_1\\ g_2 \\g_3 \end{array}\right]}.$$

The “row by column” rule of multiplication of matrices conveniently emphasises the fundamental fact that the row vector of fruits and the column vector of prices belong to different vector spaces. For example, they cannot be added — you do not add fruits and prices.

The “row by column rule” is a convenient symbolical expression of a construction know in linear algebra as paring of vector spaces, see Dual pair – Wikipedia. Actually, in this fruit bowl example, it would be natural to take one step further and use upper and lower indices in the notation,

$$p_1g^1 + p_2 g^2 +p_3g^3 = \left[\begin{array}{ccc} p_1 & p_2 & p_3\end{array}\right]\cdot \left[\begin{array}{c} g^1\\ g^2 \\g^3 \end{array}\right].$$

There are deep algebraic reasons for writing matrix products the way this is done, but I do not wish to go into pretty abstract stuff and prefer to limit myself to the most elementary justification known to me.

03/15/19

# Why do people have to learn algebra?

My answer to a question on Quora: Why do people have to learn algebra? You have no use for it.?

It could be argued indeed that it is wrong that everyone is forced to learn algebra at school.

However, without (pretty basic, between us) school algebra further study mathematics, or statistics, or computer programming is impossible.

Not learning algebra dramatically narrows further educational choices. Catching up later is very difficult — if you have not learnt algebra as a child or, at the latest, in your teen years, to do this later is of course possible but requires a degree of determination and self-discipline not normally found in general population.

Therefore, without algebra, the education system is less democratic. This is already a heavy price to pay.

In general, mathematical education is increasingly the issue of democracy.

Proper mastery of mathematics gives a person the ability to create in the head mental images of complex systems and operate with them, see them. We are surrounded by complex systems, we drown in them — in stuff starting from smartphones to the Internet and social media . Very complex computer programs make more and more decisions on peoples’ lives – whether someone can be hired to a particular job, or given a bank credit, or sold a a particular insurance policy, and so on.

Properly mathematically educated people are sighted among the blind.

If you wish to stay blind – it is your choice. Or you may think that it is entirely your choice. Actually it could happen that it is already pre-determined by a dismally bad education which, most likely, had been offered to you so far — otherwise, perhaps, you would not ask the question “ Why do people have to learn algebra?”.

03/14/19

# What do you think about opinion that Mathematics is mother of all sciences?

My answer to question on Quora: What do you think about opinion that Mathematics is mother of all sciences?

I doubt that mathematics is mother of psychology. Or of astronomy; I share a view of those historians of mathematics who suggested that it was astronomy that was the mother of mathematics. Or of physics.

I am prepared to treat with respect — even if I disagree — rather extreme views on the place of mathematics in human civilisation. I quote, for example, from Vladimir_Arnold’s paper On teaching mathematics:

Mathematics is a part of physics. Physics is an experimental science, a part of natural science. Mathematics is the part of physics where experiments are cheap.

The Jacobi identity (which forces the heights of a triangle to cross at one point) is an experimental fact in the same way as that the Earth is round (that is, homeomorphic to a ball). But it can be discovered with less expense.

Indeed let us turn to astronomy and make a thought experiment. Imagine that in the last ten thousand years the climate on the Earth was slightly different: with the same temperature an precipitation, but with a mist in atmosphere which did not decreased the luminosity of sunlight (so that agriculture is not affected), but scattered light, turning Sun and Moon into diffused light blobs and completely hiding the stars. Would mathematics have a chance to develop? For millennia, the starry sky was the only source and standard of precision — and, until relatively recent times when more precise periodic processes were discovered (thanks to physics) — absolute precision). Would mathematics develop beyond basic arithmetic and primitive geometry for agricultural use? Would the concept of mathematical rigour emerge?

In my opinion, very complex interactions between sciences do not fit into a crude “mother-daughter” simile.

03/14/19

# Is there any beginner friendly books about algebra?

My answer to a question on Quora: Is there any beginner friendly books about algebra?

The answer depends on the level at which you wish to study algebra, and on your background. So I making two assumptions. The first one is

• You wish to study abstract algebra at undergraduate level.

My second assumption involves a dichotomy. If you learn with an established education system and follow an established curricular path, then stick to textbooks which are normally used by colleges /universities in your country at the next stage of education.

If you are a self-learner or an advanced learner who is working ahead of curriculum, then I suggest to consider a truly classical book,

G. Bikhoff and S. Mac Lane, A Survey of Modern Algebra.

Why? On the first page of Preface they say:

We have tried throughout to express the conceptual background of the various definitions used. We have done this by illustrating each new term by as many familiar examples as possible. This seems especially important in an elementary text because it serves to emphasize the fact that the abstract concepts all arise from the analysis of concrete situations.

They start the book with the most classical of all algebraic structures: the ring of integers and use it to introduce commutative integral domains. They do that long before they introduce groups. In my opinion, this is a right approach, to start with something very familiar. However, here is the catch: look at this paragraph from page 3:

They take for granted that the reader understands understands without further explanation this sentence:

In $$\mathbb{Z}[\sqrt{2}], \quad a+b\sqrt{2} = c+d\sqrt{2}$$ if and only if $$a=c, \quad b=d$$.

Is it familiar to you? If you can recognise in these words a classical and very old fact of mathematics (frequently mentioned in secondary school mathematics) than the book is perhaps for you.

So if you are a self-learner of mathematics (for example, if you are in a college or university where teaching is below your level) then you also have to take responsibility for gauging the level of your readiness for study of something new.

03/14/19

# Why does algebra have letters in sums?

My answer to a question in Quora: Why does algebra have letters in sums?

One should not underestimate the influence of François Viète who was the first to use algebraic notation (letters) not only for unknowns but also for parameters (knowns) in a problem. He also used, as Wiki states, “simplification of equations by the substitution of new quantities having a certain connection with the primitive unknown quantities”. Importantly, Viète is the first cryptographer and cryptanalist known to us by name. His decryption of intercepted diplomatic correspondence had direct effect on European politics of his time. A really juicy bit from the Wiki:

In 1590, Viète discovered the key to a Spanish cipher, consisting of more than 500 characters, and this meant that all dispatches in that language which fell into the hands of the French could be easily read.

Henry IV published a letter from Commander Moreo to the king of Spain. The contents of this letter, read by Viète, revealed that the head of the League in France, the Duke of Mayenne, planned to become king in place of Henry IV. This publication led to the settlement of the Wars of Religion. The king of Spain accused Viète of having used magical powers

At that time, encryption of texts mostly used substitution ciphers, and the idea of substitution of letters for numbers should be very natural for Vieta.

I modestly suggest that teachers could perhaps use this idea: teaching primary school children some basic substitution ciphers: it is fun, it is a natural spelling exercise, and, I believe, a good propaedeutic for later study of algebra and computer coding.

03/14/19

# Why do many physicists disparage mathematical rigour?

My answer to a question on Quora: Why do many physicists disparage mathematical rigour?

Perhaps “disparage” is a wrong word, as it has already been pointed out in this discussion. However, the question is well pointed. My answer will be short and excessively general:

In mathematics, “rigour” is a tool for, and source of, consistency of mathematical results: they remain the same no matter how they are proven, results of calculation do not depend on the way the calculation was carried out, etc.

In physics, the nature itself, the experiment is a source of consistency. The world around us is remarkably consistent.

Perhaps the same can be said with smaller words: mathematicians build the ideal world, and rigour is a way to ensure that it does not collapse. Physicists have a world ready for them, and so far it does not show any sign of collapse.

03/7/19

# What number is between 1/2 and 8/9?

My answer to a question on Quora: What number is between 1/2 and 8/9?

John K Williamsson gave a good answer: for what he called the “dirty sum” of $$\frac{1}{2}$$ and $$\frac{8}{9}$$ (the mathematical term for that is mediant):

$$\displaystyle{\frac{1}{2} < \frac{1+8}{2+9} < \frac{8}{9}}$$

He suggests to use algebra to prove the mediant inequality: if $$a,b,c,d$$ are positive numbers and

$$\displaystyle{\frac{a}{b} < \frac{c}{d}}$$

then

$$\displaystyle{\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}}.$$

I wish to add that, at the primary school level, the mediant inequality does not need an algebraic proof, it is sufficiently self-evident.

Indeed consider fractions $$\frac{1}{2}$$ and $$\frac{8}{9}$$ as descriptions of real-life situations:

$$\frac{1}{2}$$ : $$2$$ children have $$1$$ bag of fruits.

$$\frac{8}{9}$$ : $$9$$ children have $$8$$ bags of fruits.

They come together and share equally: $$1 + 8$$ bags of fruit between $$2+9 = 11$$ kids, that is, they form the mediant:

$$\displaystyle{\frac{1+8}{2+9}}$$

In this sharing, which group of kids looses and and which one gains? Of course $$2$$ children with $$1$$ bag gain: they have $$\frac{1}{2}$$ bags per head, the other group comes with bigger share per head: $$\frac{8}{9}$$. For the same reason, kids in the second group lose.

I use an example with kids and bags of sweets in my lectures; here I replaced sweets by more politically correct fruits — perhaps I have to go further and use green vegetables in place of fruits. The original idea belonged to the great Israel Gelfand, and was stated in a more colorful language:

You can explain mathematics to everyone, even to drunkards. If you ask some people drinking vodka on a park bench, what is is bigger, $$\frac{2}{3}$$ or $$\frac{3}{4}$$, they will respond with expletives. But if you ask them, what is better, $$2$$ bottles of vodka for $$3$$ people or $$3$$ bottles of vodka for $$4$$ people, they will instantly give you the right answer: of course, $$3$$ bottles for $$4$$ people.

And this instant conclusion comes from an argument which is the reversal of the informal proof of the mediant inequality: how to get from the situation “$$2$$ bottles for $$3$$ people” to the situation “$$3$$ bottles for $$4$$ people”? Of course, it means that a fourth man comes and brings with him a whole bottle — can you imagine, a whole bottle of vodka! In the mediant inequality,

$$\displaystyle{\frac{2}{3} < \frac{2+1}{3+1} < \frac{1}{1}},$$

or

$$\displaystyle{ \frac{2}{3} < \frac{3}{4} < 1.}$$

I have seen some papers which confirm that this is a typical pattern of arithmetical thinking, as done by “normal” people in real life situations (for example, I have seen a claim that it is used by hospital nurses for comparing doses of medication, which one is bigger and which one is smaller).

Later addition: there are several version’s of Gelfand’s story floating on the Internet, this is one, from “Love and Math: The Heart of Hidden Reality” by Edward Frenkel, is interesting because it triggers the recursive use of the mediant inequality:

“People think they don’t understand math, but it’s all about how you explain it to them. If you ask a drunkard what number is larger, 2/3 or 3/5, he won’t be able to tell you. But if you rephrase the question: what is better, 2 bottles of vodka for 3 people or 3 bottles of vodka for 5 people, he will tell you right away: 2 bottles for 3 people, of course.”