Question

If an n × n matrix A has all the n eigenvalues λ1, λ2, ..., λn,...

If an n × n matrix A has all the n eigenvalues λ1, λ2, ..., λn, prove that det(A) = λ1 · λ2 · ... · λn.

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Answer #1

Given that A is a   matrix which has n eigenvalues .

Therefore, we know that the characteristics polynomial of A is

.

Again, given that are roots of the characteristics equation P(t) = 0 and P(t) is polynomial of degree n. So we can write P(t) as follows

.This implies that

Now we substitute t = 0 in the above equation, we have

Hence we can say that

. Hence proved.

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