Minimal Polynomial

**azdang** · March 25th 2009, 07:27 AM

Here is my problem:

Let A be a complex n x n matrix with minimal polynomial q(x)=the sum from j=0 to m of $\alpha_j x^j$ where $m\leq n$ and $\alpha_m$ = 1.

Show: If A is non-singular then $\alpha_0$ does not equal 0.

So, I get that 0=q(A)= $\alpha_0 I_n + \alpha_1 A + \alpha_2 A^2 +...+A^m$ , but I'm not sure what to do here. I assume we will have to use the fact that A is non-singular, but I'm not sure how. Does it maybe involve multiplying both sides by x on the right side? Any hints would be much appreciated!

**siclar** · March 25th 2009, 10:30 AM

If $\alpha_0$ is zero, then $0=\alpha_1 A + \alpha_2 A^2 +...+A^m=A(\alpha_1 I + \alpha_2 A +...+A^{m-1})<br />$ . But since $A$ is invertible...

**azdang** · March 27th 2009, 07:40 AM

I have tried before to assume that $\alpha_0$ is 0, but I can't figure out what to do after. Because A is invertible, we can multiply both sides by $A^{-1}$ so we get:

0= $\alpha_1+\alpha_2 A+...+A^{m-1}$ .

Buth I haven't been able to understand how that poses a contradiction.

Does it imply that all the alphas would have to be 0? Which would be aproblem because $\alpha_m$ was given to be equal to 1.

**azdang** · March 27th 2009, 07:50 AM

Oh!!! Does 0=

imply that the there would have to be a polynomial of lower degree (m-1)? Which would be impossible because we were talking about the minimal polynomial to begin with.

**ThePerfectHacker** · March 27th 2009, 11:40 AM

Originally Posted by azdang

Oh!!! Does 0=

imply that the there would have to be a polynomial of lower degree (m-1)? Which would be impossible because we were talking about the minimal polynomial to begin with.

Yes.

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