# Characteristic Polynomial

1. Apr 14, 2006

"Let $$m_T(x), f_T(x)$$ denote the minimal and characteristic polynomials of T, respectively. Let k be a scalar. Show that

$$m_{T-k}(x) = m_T(x+k)$$ and $$f_{T-k}(x)=f_T(x+k)$$."

I was able to show that the minimal polynomials were the same. But my argument was based on the minimality of the degree of m_T and it fails for characteristic polynomials.

2. Apr 15, 2006

### AKG

fT-kI(x) = det((T-kI) - xI) = det(T - (x+k)I) = fT(x+k), n'est ce pas?

3. Apr 15, 2006

I don't know. You tell me. We haven't defined what a determinant is.

4. Apr 15, 2006

### Hurkyl

Staff Emeritus
I'm curious how you define characteristic polynomial, then. AFAIK, most texts do it in terms of the determinant.

(Incidentally, it turns out that the determinant is the constant term of the characteristic polynomial)

5. Apr 15, 2006

Suppose $$d_i$$ is the dimension of the eigenspace corresponding to $$\lambda_i$$, then the characteristic polynomial, denoted $$f_T(x)$$, is:

$$f_T(x)=(x-\lambda_1)^{d_1}...(x-\lambda_r)^{d_r}$$.

Of course the two definitions must be equal, however since we haven't 'seen' the other one, I'm supposing that I can't use this equivalence to solve the problem.

6. Apr 15, 2006

### matt grime

That cannot be the definition of characteristic polynomial. The characteristic poly of an nxn matrix is a deg n poly. There is nothing that asserts that the number of eigenvalues of a matrix counted with multiplicity is n.

Eg. the 2x2 matrix with row 1 (1,1) and row 2 (0,1) has characteristic polynomial (x-1)^2, and minimal poly (x-1)^2 too, yet the dimension of the 1-eigenspace is 1.

7. Apr 15, 2006

Well I gave what I have here. I think it would work out if we replaced d with $$d_i = \dim null (T-\lambda_i I)^{dim V}$$.

8. Apr 15, 2006

### Hurkyl

Staff Emeritus
What are the eigenvalues and eigenvectors of T + kI?

9. Apr 15, 2006

If e is an eigenvalue of T, then e-k is an eigenvalue of T+kI.

10. Apr 15, 2006

### Hurkyl

Staff Emeritus
That doesn't look right. What if T = 0 and k = 1?

11. Apr 15, 2006

### AKG

I think you're assuming the underlying field to be closed, and I think you intend the $d_i$ to be the dimensions of the generalized eigenspaces corresponding to $\lambda _i$, which is equal to what you've said in post 7.

12. Apr 15, 2006

### HallsofIvy

Staff Emeritus
If v is an eigenvector of T with eigenvalue $\lambda$ then
(T+ kI)v= Tv+ kIv= $\lambda v+ kv$=$(\lambda+ k)v$.
You have a sign wrong.