Show Non-Singular Matrix A Has Non-Zero Minimal Polynomial Coefficient

[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! :)

 

[Office_Shredder]

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

 

[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!

 

[Office_Shredder]

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

 

[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

 

[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.

 

[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.

 

[xepma]

Yea, that last statement sounds about right.