Targil 9.
This targil is about eigenvalues.
1. Is it true for any real square matrix A that it is similar to AT ?
(in other words, is it true that for any A there’s invertible Q such that AT = QAQ-1).
Solution. Let us J = PAP-1 be the Jordan form. The transposition of A is similar to
JT, by the means of (PT)-1. Every Jordan block is similar to its transpose, by the
means of a permutation matrix (ones on the secondary diagonal, zeros elsewhere).
Hence J is similar to JT, and A is similar to J, and AT to JT.
Therefore A is similar to AT.
2. A is real skew-symmetric matrix (meaning A = – AT).
a. Prove that its eigenvalues are imaginary,
b. Can it have non-trivial Jordan cells (of size > 1)?
Solution. a. Assume Av  v . By * we shall denote the composition of
transposition and complex conjugation. So if v is a column vector, v* is a row of
complex conjugates to its coordinates.
Then  v  v*v  v*  Av    v* A v    v* A*  v    Av  v  v*v   v .
2
*
2
Hence, if v is nonzero vector,    . Hence  is imaginary.
b. No. Matrix iA is Hermitian, hence it can be diagonalized by a unitary matrix.
Matrix A can be diagonalized by the same unitary matrix.
3. On a circle, n real numbers are written, not all of them are equal. Each second,
all numbers change simultaneously: every number is replaced by the average of
itself and its clockwise neighbor.
a. Prove that all numbers on the circle converge to the average of all numbers.
b. Prove it converges exponentially, in other words, the distance of a value at some
point after k steps from its limit value is O(αk) where is a real number in (0, 1).
c. Compute the value of that α (the minimal value for which that claim will hold
for any starting configuration).
Solution. Let C be the matrix of cyclic permutation. It has ones on the diagonal
above the main diagonal, and one in bottom-left corner, and zeros elsewhere.
1 
 
n
Its eigenvalues are roots of x – 1, and its eigenvectors are multiples of  2  , in
 
 
 ... 
other words a geometric progression, with ratio  which is a root of xn – 1.
We can diagonalize the matrix by choosing the above vectors as a basis (in fact,
switching to this eigenbasis is so often convenient, that it has a special name –
discrete Fourier transform).
So, our operator could be written as A = (1 + C)/2, where 1 is the unit matrix.
In the eigenbasis of C, it becomes diagonal. On the diagonal we have the numbers
1 
1 
where  n  1. Since  is on the unit circle,
are on the circle whose
2
2
diameter is the interval [0, ½].
Therefore, the eigenvalue of A which corresponds to the constant vector is 1, and
all other eigenvalues are less than one. We shall denote vector ( 1, 1, …. 1) by v1.
Easy exercise sum of coordinates of any other eigenvector is zeroes.
So, assume we represent a given vector as a linear combination of eigenbasis. The
coefficient of the v1 is the average of the coordinates of the original vector (that
follows easily from the above easy exercise). When we multiply it by A many
times, this coefficient of v1 remains always the same, and all other coefficients are
reduced in geometric progression.
The two geometric progression that reduce in the slowest rate are those that
correspond to eigenvalues of highest norm, and these are closest to 1, those are

1,2 

1 e
2
1 e
2
2 i
n
2 i
n
. The absolute values of these two are
i

e
n
e
2

i
n
e

i
n
i
  
 
 cos    e n  cos   .
n
n
So, the deviation from the limit is s1k  t2k + geometrical progressions which

 
descend more rapidly. Therefore, the convergence is O  cos    . The
k

n 
progressions rotate around the circle with a step of full circle/n, so in each n
consequent steps there is a step when they don’t cancel out.
Of course, for some initial states geometric sequence can converge more rapid:
these correspond to the cases when the coefficients which correspond to these two
eigenvectors are zero, for instance when the sequence has a sub-period.
4. Prove: if the minimal polynomial of A is  x  1  1  x  2  2  ...   x  k  k , then
n
n
n
A I 
n 1
n 1
n 1
the minimal polynomial of 
is  x  1  1  x  2  2  ...   x  k  k .

 0 A
Solution.
n
 A I   An nAn1 

 0 A  
An 

  0
This follows by induction: n = 1 is obvious, and
n 1
A I 
 A I   An nAn1   An1  n  1 An 

.
 0 A   0 A
n 
n 1
0
A
0
A




 

  A I    p  A p '  A 
Therefore for any polynomial, p  
 . Hence p has to be
   0
p
A
0
A


 


n
n1
n2
such that both p and p' are divisible by  x  1   x  2   ...   x  k  k . Therefore
i is a root of degree ni + 1 at least, and vice versa – if all each i is a root of
degree ni + 1, we get zero. So  x  1  1
polynomial.
n 1
 x  2 
n2 1
 ...   x  k  k
n 1
is the minimal
5. Denote A# = (I + A)-1(I – A).
a. Show that A## = A.
b. Show that A is an orthogonal matrix iff A# is skew-symmetric.
Solution. a. First, notice that A# is defined only if –1 is not an eigenvalue of A.
It doesn't create a problem:
A# = (I + A)-1(I – A) = (I + A)-1(2I – (I + A)) = 2(I + A)-1 – I
So, if A# is defined, then A# is invertible matrix minus unit matrix, hence it doesn't
have –1 as an eigenvalue. So, if we can define A# we can also define A##.
This condition can also be rewritten as (I + A#) = 2(I + A)-1 which is the same as
(I + A#)(I + A) = 2I
Which is certainly symmetric, hence A##=A.
b. We shall use the equation (I + A#)(I + A) = 2I.
If A is orthogonal, which means ATA = AAT = I, let us transpose the previous
equation:
(I + AT)(I + A#T) = 2I
If product of two matrixes is I, then they commute (since they are mutually
inverse). Hence if their product is 2I, they commute anyway. Therefore:
(I + A#T)(I + AT) = 2I
Multiply both sides by A from the right:
(I + A#T)(A + I) = 2A
Sum it with the original equation, (I + A#)(I + A) = 2I. We get:
(2I + A# + A#T)(I + A) = 2(I + A)
We live under assumption that -1 is not an eigenvalue of A, that is, I + A is
invertible. Hence 2I + A# + A#T = 2I. That is, A# + A#T = 0, QED.
The other direction: assume A + AT = 0.
(I + A)A# = I – A (*)
Transpose :
A#T(I + AT) = I – AT
A#T(I – A) = I + A
We knew that I + A is invertible, but now we see that I – A is invertible (they are
invertible anyway, since eigenvalues of A are imaginary as we saw in problem 2a
above). Multiply the last equality by (*) from the left
(I + A)A#A#T(I – A) = (I – A)(I + A)
Obviously I + A and I – A commute, and we saw they are invertible, hence we can
cancel them out, and we get A#A#T = I.
QED.
Remark. This transformation is called Cayley transform. For complex numbers,
Cayley transform is a special Möbius transformation, which is
1 z
which moves
1 z
imaginary axis to the unit circle (and vice versa), thus transforming a circle to a
halfplane. It is not surprising that the same transformation for matrices
interchanges orthogonal matrixes (whose eigenvalues live on the unit circle) with
anti-symmetric (whose eigenvalues live on the imaginary axis).