# Does Transpose preserve eigenvalues over the complex field?

I understand from this answer that a Matrix A and its transpose $A^T$ share the same eigenvalues. This is because the transpose operation is commutative, transpose does not change the determinant and we obtain the characteristic polynomial via the determinant.

But I was solving a problem from the ECE GATE 2014 and it seems that A needs to have only real values in order for the eigenvalues to be the same. Since the answer is (b) as eigenvalues of a symmetric matrix can be negative.

I tried experimenting with different matrices populated with complex values and I find that the eigenvalues are the same.

Right now I am guessing that when we consider all complex matrices, $A^T$ means A conjugate transpose and not simply transpose. So in the question, they meant $\bar A^T$. In which case,

$$\det(A^{T} - \lambda I) = \det((A - \bar \lambda I)^{T}) = \det (A - \bar \lambda I)$$

So my understanding is that the transpose operation causes the eigenvalues to be conjugated.

My examples also show the same result.

So what is the answer to the GATE question no. 93? Why is (c) not the answer?

• Since $\det A = \det A^T$, you see that $\det (\lambda I -A) = \det (\lambda I -A^T)$. Hence the eigenvalues are the same. No need for $A$ to be real. Commented Jul 26, 2018 at 17:14

Because statement (c) is true, as well as statements (a) and (d). Only (b) is false (take $-\operatorname{Id}$, for instance).
• I mean, why does A have to be real. Why can't A be complex and still $A^T$ seems to have the same eigenvalues. Commented Jul 26, 2018 at 16:48
• So if (c) was "If A is complex, the eigen values of A and $A^T$ are always the same", even then (c) would be true? Commented Jul 26, 2018 at 16:51
• Yes. And if, instead of complex matrices, we were working with matrices over the field of rational functions in $10$ variables over $\mathbb C$, it would still be true. Commented Jul 26, 2018 at 16:54