1. ## Minimal Polynomial

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 $\displaystyle \alpha_j x^j$ where $\displaystyle m\leq n$ and $\displaystyle \alpha_m$ = 1.

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

So, I get that 0=q(A)=$\displaystyle \alpha_0 I_n + \alpha_1 A + \alpha_2 A^2 +...+A^m$, 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!

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

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

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

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 $\displaystyle \alpha_m$ was given to be equal to 1.

4. 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.

5. 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.