By Felix R. Gantmacher
This treatise, through one in every of Russia's top mathematicians, supplies in simply obtainable shape a coherent account of matrix concept so one can purposes in arithmetic, theoretical physics, statistics, electric engineering, and so forth. the person chapters were saved so far as attainable autonomous of one another, in order that the reader accustomed to the contents of bankruptcy 1 can continue instantly to the chapters of precise curiosity. a lot of the fabric has been to be had earlier in simple terms within the periodical literature.
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Additional info for The Theory of Matrices [Vol 2]
Every triangle can be cut into any number n of triangles similar to each other, except the first three primes: 2, 3, and 5. ⊓ ⊔ I created Grand II in April 1970, when I served as one of the judges of the Soviet Union National Mathematical Olympiad. The judges liked the problem. They selected the critical part of it for the juniors (ninth graders) competition: Can every triangle be cut into five triangles similar to each other? Then came the meeting to approve the problems with the Chairman of the Organizing Committee, Andrej Nikolaevich Kolmogorov, one of the greatest mathematicians of the twentieth century.
2 Excursion in Linear Algebra 31 How do we find characteristic values λ and the corresponding x a11 a12 a13 → characteristic vectors v = y of a matrix A = a21 a22 a23 ? z a31 a32 a33 No problem: according to (16), a11 a12 a13 x x a21 a22 a23 · y = λ y . z z a31 a32 a33 By performing the indicated multiplications in the matrix equation above and equating the corresponding components of the left and right sides, we get the following system of equations: a11 x + a12 y + a13 z = λx a21 x + a22 y + a23 z = λy a31 x + a32 y + a33 z = λz or, equivalently, ( a11 − λ) x + a12 y + a13 z = 0 a21 x + ( a22 − λ)y + a23 z = 0 (17) a31 x + a32 y + ( a33 − λ)z = 0.
8, then it must have been cut into two triangles by one of its diagonals. 8) that is split into two triangles similar to each other. From here we travel to a contradiction by exactly the same way as in the previous case (when the middle piece was a triangle). You, my reader, may think that Grand II is solved. In fact we “only” proved that a triangle T with integrally independent angles can not be cut into 2, 3, or 5 triangles similar to each other. But what if such a triangle T does not exist? Cheer up: it exists!