Let me first state the definition

Let $A,B\in\mathbb{C}^{n\times n}$. If there exists a nonsingular matrix $S$ s.t.

(a) $B=SAS^*$, then $B$ is said to be *-congruent or conjuctive to $A$.

(b) $B=SAS^T$, then $B$ is said to be T-congruent or congruent to $A$.

It is easy to see that both *-congruence and T-congruence are equivalence relations. Also, the triplet $(i_{+}(A),i_{-}(A),i_0(A))$ in which $i_{+}(A)$, $i_{-}(A)$ and $i_{0}(A)$ is the number of positive, negative and zero eigenvalues of $A$, respectively, is called the inertia of $A$. Also, $\mathrm{rank}(A)=i_{+}(A)+i_{-}(A)$.

The question I have is the following:

How many disjoint equivalent classes under *-congruence are there in the set of $n\times n$ complex Hermitian matrices? In the set of $n\times n$ real symmetric matrices?

Sylvester's Theorem (4.5.8 in Horn and Johnson's Matrix Analysis book) says

Hermitian matrices $A,B\in\mathbb{C}^{n\times n}$ are *-congruent if and only if they have the same inertia.

Based on this, for the 1st part of the question I tried to count cases for $(i_{+}(A),i_{-}(A),i_0(A))$, for a Hermitian matrix $A$ as a representative of its equivalence class. So for $i_0(A)=0$, the possible values of $i_{+}(A)$ are $i_{+}(A)=(0,1,\dots,n)$ and the corresponding values of $i_{-}(A)$ are $i_{-}(A)=(n,n-1,\dots,0)$. I did the same thing for all values of $i_0(A)=0,1,\dots,n$ and then summing all the possibilities, the number of disjoint classes we can have is $$(n+1)+n+\dots+1=\frac{(n+2)(n+1)}{2}$$

At this point, I will cite Thm.4.5.12 which says

Symmetric matrices $A,B\in\mathbb{C}^{n\times n}$ are T-congruent if and only if they have the same rank.

Hence, for the 2nd part of the questions I took all cases for $i_0(A)=0,1,\dots,n$ and for each of them we are only interested in the sum $i_{+}(A)+i_{-}(A)=\mathrm{rank}(A)$ and not each of them independently. So the number of disjoint classes in this case is only $$0+1+\dots+n=n+1$$ Are the above results correct?


Yes, your results are correct. Notably, you could answer that first part more efficiently by using the (second) "stars and bars" formula.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.