Say I have a linear map $T \in L(V)$. There can be multiple bases of $V$ and hence multiple matrices which represent the transformation $T$ in the form of a matrix, if we think of matrices with respect to ordered bases. So it seems to me right now that there can be multiple characteristic polynomials, each corresponding to a particular matrix representation of $T$? And solving each of them should give me the same roots i.e. eigen values? Is this true?
Is the characteristic polynomial of a linear transformation T unique
linear algebralinear-transformations
Related Solutions
Let $T$ be an invertible linear transformation from an $n$-dimensional vector space to another, $A$ its $(n \times n)$ matrix, $\mathcal B = \{\alpha_1, \dots, \alpha_n\}$ a basis for the domain, $\mathcal B' = \{\beta_1, \dots, \beta_n\}$ a basis for the co-domain, and $\beta = y_1\beta_1 + \cdots + y_n\beta_n$ a vector in the co-domain.
The inverse of $T, T^{-1},$ is related to $A^{-1}$ by $\left[T^{-1}\beta \right]_{\mathcal B} = A^{-1}[\beta]_{\mathcal B'},$ where $[ \cdot ]_{\mathcal B}$ denotes the coordinate matrix relative to the ordered basis $\mathcal B.$ Then with $A_{ij}^{-1}$ representing the $ij$ entry of $A^{-1},$ we compute $$\begin{align} \left[T^{-1}\beta \right]_{\mathcal B} & = A^{-1}[\beta]_{\mathcal B'}\\\\ & = A^{-1}[y_1\beta_1 + \cdots + y_n\beta_n]_{\mathcal B'}\\\\ & = A^{-1}\begin{bmatrix}y_1\\ \vdots\\ y_n\end{bmatrix}\\\\ & = \begin{bmatrix}\sum_{i=1}^nA_{1i}^{-1}y_i\\ \vdots\\ \sum_{i=1}^nA_{ni}^{-1}y_i\end{bmatrix};\\\\ T^{-1}\beta & = \alpha_1 \sum_{i=1}^nA_{1i}^{-1}y_i + \cdots + \alpha_n \sum_{i=1}^nA_{ni}^{-1}y_i.\end{align}$$
If you are satisfied having your inverse in terms of $\mathcal B,$ we are done.
If, however, you want the inverse in terms of the standard ordered basis, express each $\alpha_i$ in terms of the standard basis, and group the terms in the last expression above by the elements of the standard basis.
For an example, see my answer to Find linear transformation given its matrix representation. The procedure there is the same as that of obtaining $T^{-1}$ from $A^{-1}.$
No. if two matrices $A$ and $B$ represent the same linear transformation, then they are related by conjugation, $B=P^{-1}AP$, where $P$ is the change of basis. Such matrices are called similar. Two matrices may have same determinant but not be similar. They are similar if and only if they have the same canonical form.
For example
$$\left(\begin{matrix} 1 & 0\\ 0 & 1\end{matrix}\right)$$
and
$$\left(\begin{matrix} 1 & 1\\ 0 & 1\end{matrix}\right)$$
Both have determinant 1, but they are not similar.
Best Answer
A change of basis corresponds to a conjugation by an invertible matrix (i.e. we consider $PAP^{-1}$ instead of $A$). However, $$\det(PAP^{-1}-tI)=\det(PAP^{-1}-tPP^{-1})=\det(P(A-tI)P^{-1})=\\=\det (P)\det(A-tI)\det(P^{-1})=\det(PP^{-1})\det(A-tI)=\det(A-tI)$$