Proof of normal matrix criterion

  • Thread starter TTob
  • Start date
  • #1
21
0
I need to proof:
A in normal matrix if and only if [tex]trace(A^*A)=|t_1|^2+...|t_n|^2[/tex]
where [tex]t_1,...,t_n[/tex] are the characteristic roots of A.

I have a problem only with the second direction:
[tex]trace(A^*A)=|t_1|^2+...|t_n|^2[/tex] --> A is normal.

can you help me ?
 

Answers and Replies

  • #2
morphism
Science Advisor
Homework Helper
2,015
4
Presumably A is a complex matrix, so we can put it into Schur form, i.e. we can find a unitary matrix Q and an upper triangular matrix T such that A=Q-1TQ. Notice that trace(A*A)=trace(T*T). So if A is not normal, then <blank>.
 
  • #3
21
0
thanks.

note [tex]T=[\alpha_{ij}][/tex]
so [tex]trace(T^*T)=trace(A^*A)=|t_1|^2+...|t_n|^2[/tex]
hence
[tex]\sum_{\substack{
0\leq\i\leq n \\
0\leq\j\leq n
}} |\alpha_{ij}|^2
=|t_1|^2+...|t_n|^2[/tex]
because of the eigenvalues are the diagonal entries of T we have
[tex]\sum_{\substack{
0\leq\i\leq n \\
0\leq\j\leq n \\
j\ne i
}} |\alpha_{ij}|^2
=0[/tex]
hence for [tex]i\ne j[/tex] we have [tex]\alpha_{ij}=0[/tex]

so T is diagonal matrix. A=Q^-1TQ and hence A is normal.
 

Related Threads on Proof of normal matrix criterion

  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
9
Views
12K
  • Last Post
Replies
2
Views
1K
Replies
5
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
11K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
2K
Replies
8
Views
40K
  • Last Post
Replies
2
Views
3K
Top