I am trying to show that for $T$ over a complex inner product space we have $\det adj(T)=\overline{\det (T)}$.

But I have seen this, which confirms this result over the reals:


(since over the reals, $T^T=adj(T)$ and the determinants are real anyway)

But I have also seen this:

The determinant of adjugate matrix

Which seems to be very different from what I am trying to prove? I am having trouble reconciling why these two results are so different.

In fact, isnt the second result ($\det adj( T)= (\det T) ^{n-1}$) not true for say $T$ being a 3-by-3 diagonal matrix with eigenvalues $1,2,3$? (then, $\det adj( T)= \det T $)


You’re correct that the adjugate matrix isn’t the same thing as the adjoint.

To address your last point, consider $$T=\pmatrix{1&0&0\\0&2&0\\0&0&3}.$$ Its adjugate is $$\pmatrix{6&0&0\\0&3&0\\0&0&2},$$ with determinant $36$, which is indeed $6^{3-1}$.

Finally, a hint for the problem you’re trying to solve: The determinant of a matrix is equal to the product of its eigenvalues. What can you say about the eigenvalues of $T^*$ relative to those of $T$?

  • $\begingroup$ I can say that the eigenvalues of $T^*$ are the complex conjugates of $T$, but how can I say that they have the same multiplicities as well? I think those two facts would be enough. $\endgroup$
    – Andrew P.
    Nov 24 '15 at 21:42
  • $\begingroup$ The multiplicities are the same, too. Think about the characteristic polynomials of the two matrices. $\endgroup$
    – amd
    Nov 24 '15 at 22:13

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