Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\;\det(A)$?

share|improve this question
14  
Do you know that $\det(AB)=\det(A)\det(B)$? –  wj32 Apr 11 '13 at 0:48

3 Answers 3

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 $$\det(A^n) = \underbrace{\det (A) \cdot \det (A) \cdot \ldots \cdot \det (A)}_{\large n\;\text{times}} = \Big[\det(A)\Big]^n$$

share|improve this answer
    
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.

share|improve this answer
2  
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
1  
This is fantastic –  Daenerys Naharis Apr 12 '13 at 6:36
    
@Joseph Thanks for your kind words. –  user38268 Apr 12 '13 at 7:30

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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