one prove question

**wopashui** · February 24th 2010, 11:24 PM

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 $a_0 = (-1)^ndetA$

**kjchauhan** · February 25th 2010, 05:32 AM

Every square matrix satisfies its characteristic equation.

**Runty** · February 25th 2010, 05:59 AM

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.

**Bruno J.** · February 25th 2010, 06:57 AM

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 $a_0$ .

**Bruno J.** · February 25th 2010, 06:58 AM

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.

**Runty** · February 25th 2010, 10:36 AM

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.

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