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So I saw this statement in an exercise :

Two real $n \times n$ matrices are congruent if and only if they have the same rank and the same signature.

But I was wondering why do we need to state the fact that they must have the same rank. If two real $n \times n$ symmetric matrices have the same signature, doesn't they necessarily have the same rank ? So shouldn't it be :

Two real $n \times n$ matrices are congruent if and only if they have the same signature.

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  • $\begingroup$ Hmm, $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ have the same rank, but have different signatures. $\endgroup$
    – user348338
    Jun 20, 2016 at 7:25
  • $\begingroup$ Sorry, you misunderstood me. I was asking the other way round. I think that if two real $n \times n$ matrices have the same signature, then they have the same rank. But if it is so, why do we need to state it in the exercise ? It is superfluous and yet, I see it formulate like that everywhere. $\endgroup$
    – Desura
    Jun 20, 2016 at 7:28
  • $\begingroup$ Oh I see, my bad. $\endgroup$
    – user348338
    Jun 20, 2016 at 7:30
  • $\begingroup$ Sounds like Sylvester's Law of Inertia. $\endgroup$
    – almagest
    Jun 20, 2016 at 7:32
  • $\begingroup$ It probably is. It still doesn't change my question though. $\endgroup$
    – Desura
    Jun 20, 2016 at 8:36

2 Answers 2

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You are correct by any standard interpretation of the term "signature". It's difficult to know what the author is going for here.

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Some texts define the signature of a bilinear form to be the quantity $$ \text{no. of positive eigenvalues} - \text{no. of negative eigenvalues}.$$ For instance, see page 372 of Hoffman and Kunze's Linear Algebra.

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