0

If $A=P^{-1}BP$, prove $|A^k|=|B^k|$.

Please give me the idea!

Aryaman Maithani
  • 15,771
YHY 7848
  • 3
  • 1
  • What's $\lvert X\rvert$? –  Oct 13 '21 at 07:52
  • @SaucyO'Path: I'm guessing it's the order, i.e., the smallest positive $n$ such that $X^n$ is the identity matrix (and $\infty$ if no such $n$ exists). – Aryaman Maithani Oct 13 '21 at 07:56
  • 1
    @AryamanMaithani I would have guessed otherwise, and I still do. –  Oct 13 '21 at 07:56
  • If you do mean order, then look at this https://math.stackexchange.com/questions/734317/prove-that-any-conjugate-of-a-has-the-same-order-as-a – Aryaman Maithani Oct 13 '21 at 07:58
  • Oh wait, I just realised that it would make more sense for it to be interpreted as the determinant. I just haven't seen that notation in a long time. – Aryaman Maithani Oct 13 '21 at 08:00
  • I guess you mean that |A| denotes the determinant of A. If so, you can raise both LS and RS to the power k, then grouping $P^{-1}P $as $I $yields $A^k=P^{-1}B^kP $ and then taking determinant on both sides gives the result.Wish it helps. :| D – Lai Oct 13 '21 at 08:04

2 Answers2

0

Assuming that your matrices are square matrices, and $|A|$ is the determinant of the matrix $A$, you can write:

$$|A^k| = |A|^k = |P^{-1} B P|^k = (|P|^{-1} |B| |P|)^k,$$

since the determinant of a product of matrices is the product of their determinants. (Which you can prove using the explicit definition of the determinant.) The result then follows from commutativity of $\mathbb R$.

pauloss
  • 73
  • 1
  • 6
0

By induction it easy to prove that $A^k=P^{-1}B^kP$ for all $k\in\mathbb{N}$. So $|A^k|=|P^{-1}||B^k||P|$. But since $|P^{-1}|=|P|^{-1}$, $$|A^k|=|B^k|$$ for every $k\in\mathbb{N}$.

If $B$ is invertible then it can be shown that $|A^k|=|B^k|$ for every $k\in\mathbb{Z}$

Sahan Manodya
  • 982