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:

https://www.math.upenn.edu/~ekorman/teaching/det.pdf

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

But I have also seen this:

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, $\det adj( T)= \det T$)

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 $T$?
• 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. Nov 24 '15 at 21:42