Let A be an nxn matrix with characteristic polynomial

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

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

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- Feb 24th 2010, 11:24 PMwopashuione prove question
Let A be an nxn matrix with characteristic polynomial

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

Prove that $\displaystyle a_0 = (-1)^ndetA$ - Feb 25th 2010, 05:32 AMkjchauhan
Every square matrix satisfies its characteristic equation.

- Feb 25th 2010, 05:59 AMRunty
- Feb 25th 2010, 06:57 AMBruno J.
Recall that $\displaystyle f(t)=\det (A-tI)$. Set $\displaystyle t=0$ on both sides! You should just get $\displaystyle a_0$. The factor $\displaystyle (-1)^n$ could come into the picture if somehow you had defined $\displaystyle f(t)=\det (tI-A)$ but that is certainly not standard. The answer is just $\displaystyle a_0$.

- Feb 25th 2010, 06:58 AMBruno J.
- Feb 25th 2010, 10:36 AMRunty