Mnemonic for π

All the following poems describe a number.

  • 産医師異国に向こう.産後薬なく産に産婆四郎二郎死産.産婆さんに泣く.ご礼には早よ行くな.
  • Yes, I have a number.
  • How I want a drink, alcoholic of course, after the heavy lectures involving quantum mechanics.
  • Sir, I send a rhyme excelling.In sacred truth and rigid spelling. Numerical spirits elucidate, For me, the lesson's dull weight. If Nature gain, Not you complain, Tho' Dr. Johnson fulminate.
  • Que j'aime à faire apprendre un nombre utile aux sages! Immortel Archimède, artiste ingénieur, Qui de ton jugement peut priser la valeur? Pour moi, ton problème eut de pareils avantages.
  • Wie, O dies π. Macht ernstlich so vielen viele Müh!  Lernt immerhin, Jünglinge, leichte Verselein.  Wie so zum Beispiel dies dürfte zu merken sein!

The German version mentioned about the number --- π. These are all Mnemonic for π. Japanese uses sounds of numbers, but, other languages uses the number of words. From
Yes(3), I(1) have(4) a(1) number(6),
you can find 3.1416 (rounded).

I found these poems in

  • Akihiro Nozaki, A story of π, Iwanamishoten (1974) pp.77-78.

But this book also refers the followings:

  • Shin Hitotumatu, Essay for numbers, Chuuoukouronsha, (1972) p.109
  • Shigeo Nakano, the road to the modern mathematics, Shinyousya, (1973) p.19
  • Shuuichirou Yoshioka, The thousand and one nights of mathematics, Seinenshobou (1941) p.147, Gakuseisha(1959) p.107

Geometric Multiplicity: eignvectors (2)

If eigenvectors of a matrix A are independent, it is a happy property. Because the matrix A can be diagonalized with a matrix S that column vectors are eigenvectors of A. For example,

Why this is a happy property of A? Because I can find A's power easily.

A^{10} is not a big deal. Because Λ is a diagonal matrix and power of a diagonal matrix is quite simple.
A^{10} = SΛ^{10} S^{-1}
Then, why if I want to compute power of A? That is the same reason to find eigenvectors. Eigenvectors are a basis of a matrix. A matrix can be represented by a single scalar. I repeat this again. This is the happy point, a matrix becomes a scalar. What can be simpler than a scalar value.

But, this is only possible when the matrix S's columns are independent. Because S^{-1} must be exist.

Now I come back to my first question. Is the λ's multiplicity related with the number of eigenvectors? This time I found this has the name.

  • Geometric multiplicity (GM): the number of independent eigenvectors
  • Algebratic multiplicity (AM): the number of multiplicity of eigenvalues

There is no rigid relationship between them. There is only an inequality relationship GM <= AM.

For example, a 4x4 matrix's AM = 3 (The number of different λs  is 2.), GM is not necessary to be 2.

By the way, this S is a special matrix and called Hadamard matrix. I wrote a blog entry how to compute this matrix.  This matrix is so special, it is symmetric, orthogonal, and only contains 1 and -1.

The identity matrix is also an example of such matrix. The eigenvalues of 4x4 identity matrix is λ = 1,1,1,1 and eigenvectors are

I took a day to realize this. But Marc immediately pointed this out.

Though, I still think one λ value corresponds to one eigenvector in general. The number of independent eigenvector is the dimension of null space of A - λ I. The eigenvalue multiplicity is based on this as the form of characteristic function. But, I feel I need to study more to find the deep understanding of this relationship.

Anyway, an interesting thing to me is one eigenvalue can have multiple corresponding eigenvectors.

Gilbert Strang, Introduction to Linear Algebra, 4th Ed.

Geometric Multiplicity: eignvectors (1)

I had a question regarding the relationship between multiplicity of eigenvalue and eigenvectors.

I am more interested in eigenvalue's multiplicity than the value itself. Because if eigenvalue has multiplicity, the number of independent eigenvectors ``could'' decrease. My favorite property of eigen-analysis is that is a transformation to simpler basis. Here, simpler means a matrix became a scalar. I even have a problem to understand a 2x2 matrix, but a scalar has no problem, or there is no simpler thing than a scalar. Ax = λ x means the matrix A equals λ, what a great simplification!

My question is
 If λ has multiplicity, are there still independent eigenvectors for the eigenvalue?
My intuition said no. I can compute an eigenvector to a corresponding eigenvalue. But, I think I cannot compute the independent eigenvectors for one eigenvalue.

For instance, assume 2x2 matrix that has λ = 1,1, how many eigenvectors? one?

Recently I found this is related with diagonalization using eigenvector. My intuition was wrong.

For one eigenvalue, that has multiplicity, there can be multiple eigenvectors.

I will show the example of this one eigenvalue and multiple eigenventors in next article.