1. ## one prove question

Let A be an nxn matrix with characteristic polynomial

$f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1}+...+ a_1\lambda +a_0$

Prove that $a_0 = (-1)^ndetA$

2. Every square matrix satisfies its characteristic equation.

3. Originally Posted by kjchauhan
Every square matrix satisfies its characteristic equation.
That may be, but you haven't proven this. This kind of question calls for an equation proof, not just a statement.

I second the request for help on this.

4. Recall that $f(t)=\det (A-tI)$. Set $t=0$ on both sides! You should just get $a_0$. The factor $(-1)^n$ could come into the picture if somehow you had defined $f(t)=\det (tI-A)$ but that is certainly not standard. The answer is just $a_0$.

5. Originally Posted by kjchauhan
Every square matrix satisfies its characteristic equation.
I'm not sure how you expected him to use the Cayley-Hamilton theorem.

6. Originally Posted by Bruno J.
Recall that $f(t)=\det (A-tI)$. Set $t=0$ on both sides! You should just get $a_0$. The factor $(-1)^n$ could come into the picture if somehow you had defined $f(t)=\det (tI-A)$ but that is certainly not standard. The answer is just $a_0$.
EDIT: Another source told me that the term $f(t)=\det (tI-A)$ is equal to the characteristic polynomial, so that should help in solving the problem.