1
$\begingroup$

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:

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 $)

$\endgroup$
0
$\begingroup$

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 $T$?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.