Characteristic Polynomial

"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.

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

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

Hurkyl
Staff Emeritus
Gold Member
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)

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.

matt grime
Homework Helper
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.

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

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

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

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

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.
If v is an eigenvector of T with eigenvalue $\lambda$ then
(T+ kI)v= Tv+ kIv= $\lambda v+ kv$=$(\lambda+ k)v$.