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

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

1. Find spaces V*(.ABC.), V*(.AB.C.), V*(.A.BC.), V*(.AC.B.), V*(.A.B.C.) for the matrices below.

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

Answer:

To find those spaces we have:

v =  
v1
v2
v3
v4
  

Av =  
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
    
v1
v2
v3
v4
   =  
v1
v2
v3
v4
  

Bv =  
1 0 0 0
0 1 0 0
0 0 0 1
0 0 1 0
    
v1
v2
v3
v4
   =  
v1
v2
v4
v3
  

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 } = {v |  
v1
v2
v3
v4
   =  
v1
v2
v4
v3
   =  
v3
v4
v1
v2
   }

The solution is v1 = v2 = v3 = v4. The space is then generated by vector:

V*(.ABC.) = ⟨  
1
1
1
1
   ⟩

Likewise,

V(.AB.C.) = {v | Av = Bv } = {v |  
v1
v2
v3
v4
   =  
v1
v2
v4
v3
   }

The solution is v3 = v4. The space is then generated by vectors:

V(.AB.C.) = ⟨  
1
0
0
0
   ,  
0
1
0
0
   ,  
0
0
1
1
   ⟩

The strict space V*(.AB.C.) is defined as V(.AB.C.) ∩ V*(.ABC.)⊥, so we need to choose vectors orthogonal to [1 1 1 1]t and loose one dimension. One possibility is:

V*(.AB.C.) = ⟨  
0
−2
1
1
   ,  
−2
0
1
1
   ⟩

Continuing in this fashion, we have:

V(.A.BC.) = {v | Bv = Cv } = {v |  
v1
v2
v4
v3
   =  
v3
v4
v1
v2
   }

The solution is v1 = v2 = v3 = v4. The space is then equal to V(.ABC.). It follows that the strict space V*(.A.BC.) defined as V(.A.BC.) ∩ V*(.ABC.)⊥ is just {0}:

V*(.A.BC.) = {0}.

Moving on to the next space, we have:

V(.AC.B.) = {v | Av = Cv } = {v |  
v1
v2
v3
v4
   =  
v3
v4
v1
v2
   }

The solution is v1 = v3 and v2 = v4. The space is then generated by vectors:

V(.AC.B.) = ⟨  
1
0
1
0
   ,  
0
1
0
1
   ⟩

The strict space V*(.AC.B.) is defined as V(.AC.B.) ∩ V*(.ABC.)⊥, so we need to choose vectors orthogonal to [1 1 1 1]t and loose one dimension. One possibility is:

V*(.AC.B.) = ⟨  
1
−1
1
−1
   ⟩

Now, considering that with the previous strict spaces we already have dimension 1 + 2 + 0 + 1 = 4, there is no dimension left for V*(.A.B.C.), which is therefore equal to {0} as well:

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

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© 2015 Joao Meidanis