Proving detA = λ1...λn for Real Eigenvalues

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Homework Statement


Let A be nxn matrix, suppose n has real eigenvalues,λ1,...,λn, repeated according to multipilicities. Prove that detA = λ1...λn.

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The Attempt at a Solution


I started by applying the definition, Av = λv, where v is an eigenvector. then I just dun know how to keep going.. is there anyone can help me out? or at least give me some hints..
thx..
 
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If matrices A and B are similar, what can you say about their determinants? Use this relationship to help you solve the problem.
 
DH offered an correct approach,but you need to know something about similarity and Jordan normal form or Schur's lemma
Here's another approach: consider the characteristic polynomial det(A-λI),by fundamental theorem of algebra, it can be factorized into (λ1-λ)(λ2-λ)...(λn-λ),then let λ=0 and see what will happen
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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