Up to what year of mathematics research, does every mathematician need to know all mathematics discovered until that point?

My answer to a question on Quora: Up to what year of mathematics research, does every mathematician need to know all mathematics discovered until that point?

At a practical level: for a beginner mathematician, it is critically important to read classical papers in the field very carefully, line-by-line. With experience, you start scanning papers for key ideas and critical configurations where mistakes are more likely: “aha, the author uses this particular approach; it is interesting to see where and how that particular case is handled in the proof …” This works because an experienced mathematician maintains in his or her mind a mental image of his/her research field and maps the mathematical contents of the paper to that image.

It is impossible to read all publications even in a relatively narrow area of research. You have to be very selective in your reading. Life is short. Mathematics is done by the subconscious; when reading mathematics, you feed your subconscious. It is like feeding a pet — you are responsible for its well-being, and have to maintain a healthy nutritious diet.


Why are the transpose and inverse of an orthogonal matrix equal?

My answer to a question in Quora: Why are the transpose and inverse of an orthogonal matrix equal?

Orthogonal matrices have several equivalent definitions — this is reflected in previous answers to your question. It could happen that in the book that you are using, or in the lectures that you are taking, the equality of the transpose and inverse, \(A^T = A^{-1}\), is chosen as the first definition of an orthogonal matrix and its equivalence to other statements is proven later.

So my answer: depending on the chosen way of exposition of linear algebra in a book or a lecture, it could be just a definition.


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

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


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

so we can take

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


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 \(AB = 0\).


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.


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?”.


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.


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.