# Show that Characteristic polynomial = minimal polynomial

1. Mar 20, 2012

### PirateFan308

1. The problem statement, all variables and given/known data
Let A = $\begin{pmatrix}1 & 1 & 0 & 0\\-1 & -1 & 0 & 0\\-2 & -2 & 2 & 1\\ 1 & 1 & -1 & 0 \end{pmatrix}$

The characteristic polynomial is $f(x)=x^2(x-1)^2$. Show that f(x) is also the minimal polynomial of A.

Method 1: Find v having degree 4.
Method 2: Find a vector v of degree 2, whose minimal polynomial (on A of v) is $x^2$, and another, w, whose minimal polynomial is $(w-1)^2$. Or, just show that v and w exist.

3. The attempt at a solution
I'm confused as to how a vector can have a degree of more than 1. Isn't a vector simply:
$v= \begin{pmatrix}a\\b\\c\\...\\n\end{pmatrix}$ in $R^n$? I think I can get the question once I understand this. Thanks!

2. Mar 20, 2012

### Dick

I think what they mean by a vector v having degree 4 is that A^4(v)=0 but A^3(v) is not equal to zero.

3. Mar 21, 2012

### PirateFan308

Oh, ok. Thanks!

4. Mar 21, 2012

### Dick

You're welcome but rereading that I'm not sure what the hint really means. Since f is the characteristic polynomial you know f(A)=A^2(A-1)^2=0. To show it's minimal you need to show A(A-1)^2 and A^2(A-1) are not zero. The 'degree of a vector' seems pretty unclear to me as well.