# Minimal Polynomial

1. Mar 25, 2009

### azdang

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! :)

2. Mar 25, 2009

### Office_Shredder

Staff Emeritus
Assume a0 is zero. See if you can find a way to reduce the degree of the minimal polynomial such that A is still a zero

3. Mar 25, 2009

### azdang

So, first, I've tried rewriting q(A) like this:
0=$$\alpha_0 I_n+(\alpha_1+\alpha_2 A+...+A^{m-1})A$$

Then, if $$\alpha_0$$ is 0, then 0=$$(\alpha_1+\alpha_2 A+...+A^{m-1})A$$.

So, if we multiply both sides by a nonzero vector x, we have:
0=$$(\alpha_1+\alpha_2 A+...+A^{m-1})Ax$$.
But since x is nonzero and A is nonsingular, this cannot equal 0. Therefore, $$\alpha_0$$ must equal 0. Is my reasoning correct? I'm still not sure. Thanks!

4. Mar 25, 2009

### Office_Shredder

Staff Emeritus
You don't know that $$\alpha_1 + ... + A^{m-1}$$ is non-singular. Instead of applying both operators to a vector, try multiplying on the right by A-1 and see what you have

5. Mar 25, 2009

### azdang

Well, then $$\alpha_1 + \alpha_2 A+...+A^{m-1}$$=0. But I don't see how that helps because if we are assuming that $$\alpha_0$$ is 0, and then $$\alpha_1 + \alpha_2 A+...+A^{m-1}$$=0, q(A) still equals 0, so it would give us what we want. I must be missing something

6. Mar 27, 2009

### azdang

Hey guys, I'm still having problems with this. As Office_Shredder suggested, I assume $$\alpha_0$$ is 0, and then multiply by the inverse of A on the right side of each side which leaves me with:

0=$$\alpha_1 + \alpha_2 A+...+A^{m-1}$$, but what to do with this I can't seem to see.

7. Mar 27, 2009

### azdang

Oh!!! Does this last equation set equal to 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.

8. Mar 27, 2009

### xepma

Yea, that last statement sounds about right.