Misunderstanding Caley-Hamilton Theorem - Characteristic Polynomial in the Standard/Factorized Form

Question

so my question is about the Caley-Hamilton theorem. Consider the following Matrix A.

$$A =\begin{pmatrix} -1 & 0 & 4 \\ 2 & -1 & 0 \\ 3 & 2 & -1 \end{pmatrix}$$

The characteristic polynomial is (according to WolframAlpha)

$$\chi_A(X)=-X^3-3X^2+9X+27$$

or in factorized form

$$\chi_A(X)=-(X-3)(X+3)^2.$$

Now, according to CHT, $\chi_A(A)=0$. This is true in this particular case if the characteristic polynomial is in the standard form.

$$\chi_A(A)=-A^3-3A^2+9A+27I_3=0$$

But appearantly, if we use the factorized form of the polynomial, it is false.

$$\chi_A(A)=-(A-3I_3)(A+3I_3)^2\neq 0$$

So, I must be doing something wrong or misunderstanding something. I just don't get where I made a mistake. Any help would be greatly appreciated.

PS: I'm sure that I'm not the first one to ask this question, but I didn't know how to phrase it correctly to find an old answer. Also, I hope my tags are okay.

Answer 1

What you did wrong appears to have been arithmetic. Here's a transcript of a matlab session:

>> A = [-1, 0, 4; 2 -1 0; 3 2 -1];
>> eig(A)

ans =

   3.0000 + 0.0000i
  -3.0000 + 0.0000i
  -3.0000 - 0.0000i

>> I = eye(3);
>> Q= (A + 3*I)^2

Q =

    16     8    16
     8     4     8
    16     8    16

>> P = (A - 3*I);
>> P * Q

ans =

     0     0     0
     0     0     0
     0     0     0

which shows that in the factored form, the characteristic polynomial still gives zero.

Answer 2

You have to check your algebra. The same manipulations that show that $$-x^3-3x^2+9X+27=-(x-3)(x+3)^2$$ will work when writing $A$ instead of $x$.

In WA, the matrix product is denoted with a period. And the square is interpreted entrywise. Here is the computation.