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I'm trying to get ready to my exam from linear algebra by doing some random tasks and with this one i'm pretty stuck.

$$A = \begin{pmatrix} 8-5i & -6 \\ 4-5i & -3+5i \end{pmatrix}$$ Given this matrix $A$ i have to calculate determinant of $A^{-2013}$. I tried to do this by using eigenvalues. And i know this matrix has inversion ( since $det(A) \neq 0 $ ), so it's true that $det (A^{-2013}) = det (A^{2013})^{-1}$. What more i got is that eigenvalues are $\lambda_1 = 5-5i$ and $\lambda_2 = 5i$. With that, we can use similarity matrix B that $$B= \begin {pmatrix} 5-5i & 0 \\ 0 & 5i \end{pmatrix} $$ and $$A = P^{-1}\times B \times P$$ so we have to find $A \times C = B$ And this is the part i'm stuck with... I have no idea how to do it nicely, because by staight matrix multiplication i end up with 4 complicated equations. From that i think it's okay to do $ ( P^{-1} \times B \times P ) ^{2013} = (P^{-1} \times B \times P \times P^{-1} \times B \dots ) = P^{-1} \times B^{2013} \times P $, so the final answer will be $det(P^{-1} \times B^{2013} \times P)^{-1}$, since $B$ is diagonal matrix, raising it to 2013th power will be easy. So, please help me figuring out this hard part, i don't want to make it overcomplicated :-) Thanks in advance!!!

Krzysztof Lewko
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1 Answers1

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This is much simpler, $\det(A^n)=\det(A)^n$. And consider also that $(1-i)^2=-2i$.

Lutz Lehmann
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  • Oh, that's a nice trick, thanks. But i'm also stuck with Calculating the $P$ matrix. Your second hint doesn't apply to it, i think, because i have 1st equation ( from matrix multiplication ) $(8-5i)x_1 - 6x_3 = 5-5i$ and other 3 similiary complicated ones. Could you help me with this? – Krzysztof Lewko Jan 02 '14 at 22:56
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    The OP should consider the significance of the fact that $\det(\cdot)$ commutes with multiplication. This fact is enough to prove very easily that he only needs to consider powers of his diagonalized matrix, to which $\det(\cdot)$ is especially easy to apply. This means he doesn't need to know the $P$ matrix -- its determinant cancels with that of $P^{-1}$. – MPW Jan 02 '14 at 22:59
  • The equation for the eigenvector is $(8−5i)x_1−6x_2=(5−5i)x_1$ or $6x_2=3x_1$, which is satisfied for $(x_1,x_2)=(2,1)$. The other one goes in similar ways. No need to separate real and imaginary parts. – Lutz Lehmann Jan 02 '14 at 23:01
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    What I'm saying is $\det(A) = \frac{1}{\det(P)}\cdot \det(B)^n \cdot \det(P)$. – MPW Jan 02 '14 at 23:02
  • Oh... thanks a lot @MPW. Right, it makes sense now! THANKS you all – Krzysztof Lewko Jan 03 '14 at 08:46