MO640 - Exercises - Matrix medians, Zanetti, Biller, and Meidanis 2013

Exercises marked with (*) require further reading/search beyond the suggested texts.

2. Find matrix median approximation MA for:

A =  
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
  
,
B =  
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
  
,
C =  
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
  

Is it a matrix median?

Answer:

We have:

v =  
v1
v2
v3
v4
  
Av =  
0 1 0 0
1 0 0 0
0 0 0 1
0 0 1 0
  
 
v1
v2
v3
v4
  
=
 
v2
v1
v4
v3
  
Bv =  
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
  
 
v1
v2
v3
v4
  
=
 
v4
v3
v2
v1
  
Cv =  
0 0 1 0
0 0 0 1
1 0 0 0
0 1 0 0
  
 
v1
v2
v3
v4
  
=
 
v3
v4
v1
v2
  

V*(.ABC.) = {v | Av = Bv = Cv }; the solution is v1 = v2 = v3 = v4.

V*(.ABC.) = ⟨  
1
1
1
1
   ⟩

In the same way, we can find the other spaces:

V*(.AB.C.) = ⟨  
1
-1
1
-1
   ⟩
V*(.A.BC.) = ⟨  
1
1
-1
-1
   ⟩
V*(.AC.B.) = ⟨  
1
-1
-1
1
   ⟩

V*(.A.B.C.) = {0}

The median approximation MA can be found by solving:

MA  
1 1 1 1
1 -1 1 -1
1 1 -1 -1
1 -1 -1 1
  
=
 
1 -1 -1 -1
1 1 -1 1
1 -1 1 1
1 1 1 -1
  
MA =  
-0.5 0.5 0.5 0.5
0.5 -0.5 0.5 0.5
0.5 0.5 -0.5 0.5
0.5 0.5 0.5 -0.5
  

MA is a median.

d(MA;A,B,C) = d(MA,A) + d(MA,B) + d(MA,C)

d(MA;A,B,C) = 1 + 1 + 1 = 3, which is the minimum value for the total score.

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© 2015 Joao Meidanis