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How can the value of $\;\det\left(A^{11}\right)\;$ be calculated from $\;\det(A)$?

Generally how can $\;\det\left(A^n\right)\;$ be obtained from $\;\det(A)$?

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Do you know that $\det(AB)=\det(A)\det(B)$? – wj32 Apr 11 '13 at 0:48

You simply need to iteratively apply the identity that states that $$\det(AB) = \det(A)\cdot \det(B)$$ $$ \implies \det(AA) = \det (A)\cdot \det(A)\quad\quad\;\,$$ and you arrive at the fact that $$\det(A^n) = \underbrace{\det (A) \cdot \det (A) \cdot \ldots \cdot \det (A)}_{\large n\;\text{times}} = \Big[\det(A)\Big]^n$$

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Nice - I like when you can use first principles! +1 – Amzoti Apr 11 '13 at 1:24

We recall that the determinant of an endomorphism $T : V \to V$ is the unique scalar $c$ such that the functored map

$$\bigwedge\nolimits^{\!k} T : \bigwedge\nolimits^{\!k} V \to \bigwedge\nolimits^{\!k} V$$

is multiplication by $c$. Here $k$ is the dimension of $V$. Thus you are asking why if we apply $\bigwedge^k T$ $n$ number of times the resulting map is multiplication by $c^n$. But this is just obvious.

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interesting, but probably over the head of someone who seems not to recall that $det(AB) = det(A)det(B)$. – user27182 Apr 11 '13 at 19:44
This is fantastic – Daenerys Naharis Apr 12 '13 at 6:36
@Joseph Thanks for your kind words. – user38268 Apr 12 '13 at 7:30

${}{}{}{}{}{}{}{}{}{}{}det(A^n) = [det(A)]^n$.

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