Characteristic polynomial of a linear operator T Suppose that $V$ is a finite dimensional vector space with $B_1$ and $B_2$ as its ordered basis and let $T$ be a linear operator on $V$.
Then the matrices $[T]_{B_1}$ and $[T]_{B_2}$ are similar.
Why are they similar?
Also, because they are similar, it would follow that their characteristic polynomials are the same, which would further imply that, the characteristic polynomial of $T$ is independent of the choice of ordered basis, right ?
 A: For any two bases, there is a matrix $C$ such that if a vector has coordinate-vector $a$ in the first basis, then it has coordinate vector $Ca$ in the second. The best way to figure out what this matrix is is to consider the two bases as creating two entirely distinct vector spaces, $U$ for the first, and $W$ for the second.
To get the domains to match up properly, you have to have $C:U\to W$. Remembering that basis elements in $U$ and $W$ are just ones and zeros, you should be able to figure out what $C$ and $C^{-1}$ do to all vectors. It turns out that this $C$ is the right one such that $[T]_{B_2}=C[T]_{B_1}C^{-1}$.
And yes, everything you said in the last paragraph is right.
A: I will use $\beta = \{u_1, \dotsc, u_m\}$ and $\gamma = \{v_1, \dotsc, v_n\}$ as the bases. For any linear transformation $T$, we can write $T(u_j)$ uniquely in terms of $\gamma$ as $T(u_j) = \sum_{i = 1}^n a_{ij}v_i$. We can arrange $a_{ij}$ into a unique matrix and denote it $[T]_{\beta}^{\gamma}$ or just $[T]_\beta$ if both bases are $\beta$. Similarly, if $v = a_i u_i$ we can denote $[v]_\beta$ to be the column matrix of vector $v$ with respect to basis $\beta$ consisting of $a_i$.
Going through all the details, it can be proven that $[ST]_\beta^\gamma = [S]_\alpha^\gamma [T]_\beta^\alpha$ and $[Tv]_\gamma = [T]_\beta^\gamma [v]_\beta$. In particular we can consider the linear operators $T$ and $I$(identity operator) on a $n$ dimensional vector space $V$. Let $Q = [I]_\beta^\gamma$. Then, for any $v \in V$, we have
$$
[Iv]_\gamma = [I]_\beta^\gamma [v]_\beta \implies [v]_\gamma = Q[v]_\beta. \tag{1}
$$
Also,
$$
[Tv]_\gamma = [I]_\beta^\gamma [Tv]_\beta \implies [T]_\gamma [v]_\gamma = Q[T]_\beta [v]_\beta.
$$
Using $(1)$, this becomes
$$
[T]_\gamma Q [v]_\beta = Q[T]_\beta [v]_\beta \implies [T]_\gamma Q = Q[T]_\beta \implies [T]_\beta = Q^{-1} [T]_\gamma Q.
$$
Note that $Q$ is invertible since $I$ is invertible. The last equation show the transformation matrices with respect to two bases are similar.
