Thread: determinant

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

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, $\displaystyle (-1)^n\det A$ 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

PS Perhaps there is something about some free coefficient or stuff...?

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)

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 $\displaystyle p(x)=\det\left(Ix-A\right)$, right? And that the constant term of a polynomial $\displaystyle q$ is $\displaystyle q(0)$ and $\displaystyle \det\left(c A\right)=c^n\det(A)$...right?

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?

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 $\displaystyle 0$ into $\displaystyle p(x)=\det\left(Ix-A\right)$?

oh crap!!! im so blur! thanks