# one prove question

• Feb 24th 2010, 11:24 PM
wopashui
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 $a_0 = (-1)^ndetA$
• Feb 25th 2010, 05:32 AM
kjchauhan
Every square matrix satisfies its characteristic equation.
• Feb 25th 2010, 05:59 AM
Runty
Quote:

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.
• Feb 25th 2010, 06:57 AM
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 $a_0$.
• Feb 25th 2010, 06:58 AM
Bruno J.
Quote:

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.
• Feb 25th 2010, 10:36 AM
Runty
Quote:

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.