For a strictly complex matrix $A$,

1) Can we comment on determinant of $A^{*}$ (conjugate of entries of $A$) , $A^{T}$ (transpose of A) and $A^{H}$ (hermitian of $A$). I know that for real matrices, $\det(A)=\det(A^{T})$. Does it carry over to complex matrices, i.e. does $\det(A)=\det(A^{T})$ in general? I understand $\det(A)=\det(A^{H})$ (from Schur triangularization).

2) The same question as first, now about eigenvalues of $A$. I would like to know about special cases, for instance what if $A$ is hermitian or positive definite and so on.

  • $\begingroup$ this is an anchor $\endgroup$ – Ning Wang Sep 4 '18 at 14:34

Since complex conjugation satisfies $\overline{xy} = \overline{x} \cdot \overline{y}$ and $\overline{x+y} = \overline{x} + \overline{y}$, you can see with the Leibniz formula quickly that $\det[A^*] = \overline{\det[A]}$.

For complex matrices $\det[A] = \det[A^T]$ still holds and doesn't require any changes to the proof for real matrices.

Together this means that $\det[A] = \overline{\det[A^H]}$.

This applies to the eigenvalues as well: the characteristic polynomial of $A^*$ is given by $\det[tI - A^*] = \det[(\overline{t}I - A)^*] = \overline{\det[\overline{t}I - A]}$ and the eigenvalues of $A^*$ are exactly the complex conjugates of those of $A$.

In particular if $A$ is hermitian, $A = A^*$ and so all eigenvalues are equal to their complex conjugates - in other words, they're real.

  • 2
    $\begingroup$ A few more useful facts. Skew-Hermitian matrices have purely imaginary eigenvalues. Unitary matrices have eigenvalues which lie on the unit circle. Matrices with all real entries will always have eigenvalues occurring as conjugate pairs, this follows from the conjugate root theorem for real polynomials. $\endgroup$ – EuYu Oct 29 '12 at 14:24
  • $\begingroup$ @Euyu Thanks a lot both of you. I would like to explain the question which actually inspired all of this questions, I have a rank one positive definite matrix $A=xx'$. I consider $A^{T}$. Now, define $b=x^{*}$ and we have $A^{T}=bb^{H}$, by above arguments, they should have same eigen value. $\endgroup$ – dineshdileep Oct 29 '12 at 14:39
  • $\begingroup$ @dineshdileep You will have $A^T = b^* b^H$ $\endgroup$ – Cocopuffs Oct 29 '12 at 14:51
  • $\begingroup$ @Cocopuffs $A^{T}=b^{*}b^{H}=xx^{T}$, is it so? i feel something wrong with it. $\endgroup$ – dineshdileep Oct 29 '12 at 15:31

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