Thread:

1. determinant

given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?

2. Originally Posted by alexandrabel90
given that A is an nxn matrix, how do you show that the characteristic polynomial of A is (-1)^n det(A)?
Doing magic because the above is far from being true. In fact, $(-1)^n\det A$ is not even a polynomial of degree n, let alone

the characateristic polynomial of any nxn matrix. Read carefully again the question and repost.

Tonio

3. the exact question is let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the constant term of the characteristic polynomial is (-1)^n det(A)

4. Originally Posted by alexandrabel90
the exact question is let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the constant term of the characteristic polynomial is (-1)^n det(A)
You know that $p(x)=\det\left(Ix-A\right)$, right? And that the constant term of a polynomial $q$ is $q(0)$ and $\det\left(c A\right)=c^n\det(A)$...right?

5. yup, i know that..

so lets say, A is similar to an upper triangular matric so P(x) = (x-a_11)....(x-a_nn) = x^n + a_n-1 x^(n-1) +....+a_0 so P(0) = a_0 and then how do i continue?

6. Originally Posted by alexandrabel90
yup, i know that..

so lets say, A is similar to an upper triangular matric so P(x) = (x-a_11)....(x-a_nn) = x^n + a_n-1 x^(n-1) +....+a_0 so P(0) = a_0 and then how do i continue?
Friend, what if we just plugged $0$ into $p(x)=\det\left(Ix-A\right)$?

7. oh crap!!! im so blur! thanks
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