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I was wondering if there is a relation between the determinant of a matrix and the determinant of its powers. I mean I am looking for something like $$ \det (A^k) = f(\det(A), k). $$ A few check I made seem to suggest that $$ \det (A^k) = \det (A)^k $$ but I cannot find any kind of reasoning to justify such an assumption. It looks like a standard result, so I would expect it to be somewhere but I cannot find it either.

EDIT: Sorry, I forgot to specify that I'm interested in $k \in \mathbb{R}$, I noticed only when I received the first answers/comments.

gosbi
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  • Do you know that $\det (AB)=\det(A)\det(B)$? – paw88789 Jan 25 '15 at 09:07
  • @paw88789: my bad, I forgot to say where I needed t to be chosen, I'll edit the question just now. – gosbi Jan 25 '15 at 09:09
  • Maybe this answers your question – Bman72 Jan 25 '15 at 09:17
  • We're going to need you to say what is for you, for example $;A^\pi;$ . Also, using $;t;$ in this particular instance of exponent may not be a good idea as it'd be easy to confuse the exponent with the transposed matrix. – Timbuc Jan 25 '15 at 09:53

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Assume none of the eigenvalues of $A$ lie on the negative real axis $(-\infty,0]$, it is always possible to define $A^s$ in such a way that $\det(A^s) = \det(A)^s$. To see that, use a similarity transform to bring $A$ into its Jordan normal form and works with it.

If $A$ has zero as an eigenvalue, then one may get into trouble in defining $A^s$.
The simplest example I can think of is $A = \begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}$.

If all eigenvalues of $A$ are non-zero but some of them are negative, we can define $A^s$ but we may get into problem to enforce $\det(A^s) = \det(A)^s$. As an example, consider

$$A = \begin{bmatrix}-1 & 0\\ 1 & -1\end{bmatrix}$$

We have $\det(A) = 1$ but any sensible/natural definition of $A^s$ will force $\det(A^s) = e^{2\pi n s i}$ for some odd integer $n$.

achille hui
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