matrices - Characteristic polynomial of linear operator and matrix

Does the characteristic polynomial of a linear operator always equal to the characteristic polynomial of its representation under some basis $\mathcal{B}$?
That is $c_{\tau}(x) = c_{[\tau]_{\mathcal{B}}}(x)$?

Another question is why the characteristic polynomial of a symmetric linear operator may have complex root? (we know symmetric matrix always has real eigenvalues.) Is there some example?

The characteristic polynomial is defined by product of elementary divisor for $\tau$ .

The symmetric operator means $(\tau v,w) = (v,\tau w)$ that is adjoint for real bilinear form.

$[\tau]_{\mathcal{B}}$ is the matrix that columns are coordinate of $\tau(b_i)$ under basis $\mathcal{B}$

Answer

Yes: the characteristic polynomial of a linear operator always equal to the characteristic polynomial of its representation under any choice basis $\mathcal B$. The fact that this is so is a direct consequence of the structure theorem for PIDs (which presumably you have seen if you are doing anything with "elementary divisors"). In particular, there is a direct correspondence between elementary divisors of an operator and elementary divisors of the associated matrix.

For a more thorough discussion, see this wikipedia page.

Regarding your second question: if our vector space is over the field $\Bbb R$ and the bilinear form is positive definite, then "symmetric operators" will correspond to "symmetric (real) matrices", which means that the spectral theorem applies and the roots of the characteristic polynomial must be real. For other bilinear forms, we can make no such guarantee.

An example of a "symmetric" operator that fails to have real eigenvalues. Consider the bilinear form over $\Bbb R^2$ given by
$$
(x,y) = x_1y_2 + x_2 y_1.
$$
Relative to this bilinear form, we find that the operator $x \mapsto Ax$ where
$$

A = \pmatrix{0&-1\\1&0}
$$
is "symmetric", but has no real eigenvalues.