Download Targil 6. This targil is inspired by SEEMOUS 2010. 1. a.* Question

Targil 6.
This targil is inspired by SEEMOUS 2010.
1. a.* Question from the last SEEMOUS: given a real 2×2 matrix A, prove that
there are two 2×2 real matrixes B and C such that A = B 2 + C2 (this was the nicest
problem in the competition).
Solution. The nicest solution was invented by Ohad Livne (under the influence of
Minkowski) during the competition. Any matrix has the following decomposition:
 a b k m
A
0 l 

b
a

 

Where k and l are positive. The first matrix correspond to a complex numbers
when you consider it as a linear transformation of R2 . Complex number has a root
as a complex number, so that matrix is a square of a 2×2 matrix.
 k
The second matrix is a square of 
 0

x 
 , where x is easily computed:
l 
x   k x  k


l   0
l   0
m
So it is enough to take x 
. QED.
k l
 k

 0



k  l x


l

b.** The natural generalization: for real matrixes n×n, is it possible to represent
each matrix as a sum of squares, and if yes, how many squares are required?
The answer is the following:
For n = 1 it doesn’t work (there are negative numbers which are not sums of
squares).
For other odd n: three squares are required.
For even n: two squares are required.
The solution will be explained in the end.
2. a. Prove that a real matrix which is sufficiently close to the unit matrix is a
square of real matrix.
Solution. Use Newton’s binomial formula:
 
 

1  x   f  x      x n
n 0  n 
The radius of convergence for  
series converge. The identity
1
is 1. Therefore, for a matrix of norm < 1 the
2
 f  x 
12
2
 1  x can be verified directly, and it is
true for linear transformation whenever the series converge (since the powers of
the same matrix commute).
b.* Let D be a diagonal matrix with positive numbers at the diagonal. Prove that a
matrix which is sufficiently close to D is a square of real matrix. (If you don’t
know what is “sufficiently close” stop complaining and invent a definition.)
Solution. By D we shall denote the diagonal matrix, which has square roots of
D matrix elements at the respective diagonal cells.
Consider the transformation X
X 2 in the neighborhood of D . By the inverse
function theorem, the transformation is invertible in the neighborhood if the
differential is non-degenerate. Which means it is enough to verify that the
2
d
D A
 0 for each A  0 .
derivation in every direction is nonzero:
d
 0




2
d
D   A  D  A  A D .
d
In both terms of this sum, entries of A are multiplied by positive numbers (in one
The computation yields:
the rows are multiplied by respective diagonal entries of D , in another the
columns). So if A isn’t zero, the directional derivative isn’t 0. QED.
3. Is it true that every complex n×n matrix is a square of a complex matrix?
0 1
Solution. No. The matrix 
 is not a square. It is nilpotent of order 2, so the
0 0
“square root” would be nilpotent of order > 2 (which is impossible for matrixes
2×2).
4. A determinant of a 2×2 real matrix is positive. Is it true that this matrix is a
square of a real matrix?
Answer. No.
 1 0 
 B 2 (clearly, A has a positive determinant).
Solution. Assume that A  

 0 2 
Then the complex eigenvalues of B are   1,   2 so they are not conjugate to
each other hence the characteristic polynomial of B is not a polynomial with real
coefficients.
 1 0 0 
5. Is  0 1 0  a sum of two squares of real matrixes?
 0 0 1


Answer. No.
 1 0 0 
Solution. Assume C   0 1 0   A2  B 2 . The matrix C is scalar, so it has
 0 0 1


the same form in any basis. We shall choose (a complex) basis in which A is upper
triangular (for example Jordan) than also A2 is upper triangular and B2 = C – A2 is
upper triangular. The diagonal elements of A2, B2 are their eigenvalues, and they
are squares of eigenvalues of A, B. Denote a1, a2, a3 eigenvalues of A, b1, b2, b3
eigenvalues of B. There are 3 equations:
a12  b12  1
a22  b22  1
a32  b32  1
One of eigenvalues of both A and B is real, so it must be in pair with imaginary
eigenvalue of the other matrix, so that sum of the squares will be -1. But imaginary
values come in pairs, so both A and B have two imaginary values. So one of these
equations is a sum of squares of two imaginary values.
WLOG, let us assume that a1, b2 are real and others are imaginary.
Hence b12  b32  1, and a22  a32  1 , hence a32  b32  2 , which is impossible.
And, finally:
Solution for 1b.
For even case. Let m be the maximum of absolute values of the entries of A.
In ε-neighborhood of 1, square root is defined (see problem 2).
1

Take a huge number M such that 2  .
M
m
2
Write A = D + (– M 1), where 1 is the unit matrix.
Then D = M2(1 + A / M2). But 1 + A / M2 is in ε-neighborhood of 1, so it has
square root, hence D has square root.
The second summand, – M21 is a square also, because in even dimensions –1 is a
square (for n = 2 it is rotation by 90°, for greater dimension it can be constructed
from blocks). So, we have constructed decomposition into sum of two squares.
Not every matrix is a square; indeed, every negative-determinant matrix is not a
square. So 2 is the minimal number.
For odd case. We shall prove that the sum of two squares can’t give –1, therefore
in some cases at least 3 squares are needed, and construct decomposition into 3
squares for every matrix.
The construction of decomposition is similar to the even case. Let V be the
diagonal matrix such that right bottom corner is 1 and all other elements are –9.
Let U be the diagonal matrix s. t. top left corner is 1 and all other elements are –9.
These matrixes obviously have square root (take square root of –1 in dimension
less by 1, multiply it by 3, and add 1-block).
Every matrix A can be written as A = MV + MU + W, for any huge positive
number M. But if M is sufficiently huge, W/M is very close to a specific diagonal
matrix with positive numbers at the ends (8 in both corners and 18 elsewhere)
hence it will have square root by problem 2.
It remains to prove that sum of two squares can’t be –1. Assume A2 + B2 = –1,
where A, B are real matrixes. Choose a basis over complex numbers, in which A
would be upper triangular. In that basis, A2 and B2 = A2 – 1 are upper triangular.
On the diagonals we have squares of eigenvalues of A and B. That yields a system
of equations: ak2  bk2  1 .
At least one eigenvalue of A and at least one eigenvalue of B is real (polynomial of
odd degree has a real root), others are either real or complex conjugate.
Assume ak is real, ak2  0 . So bk2  1 , therefore bk is imaginary. Hence it has a
conjugate, bj. Hence a2j  0 and aj is real.
Therefore, if we fix pairing between conjugate imaginary roots of characteristic
polynomial of B, we get a pairing between real roots of A. But it is impossible,
because polynomial of odd degree has odd number of real roots (considered with
multiplicity). Contradiction, QED.