Flaw in my argument that $[T]_{B} = [T]_{B'} \implies B = B'$ I am currently self-studying linear algebra from Anton's Elementary Linear Algebra, and have come across the following question while working through the exercises: let $B$ and $B'$ be two bases for $\mathbb{R}^{n}$, and let $T:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a linear operator. If $[T]_{B} = [T]_{B'}$, is it true that $B = B'$?
This is false since the identity operator provides a counterexample for any choice of bases. However I can't reconcile how this is consistent with the following theorem, which has been stated without proof. Essentially, it says the following (though I am summarising):

Let $V$ be a finite-dimensional vector space, and suppose that the matrices $A$ and $B$ are similar with $B = P^{-1}AP$. Then $A$ and $B$ represent the same linear operator on $V$ (possibly with respect to different bases), and $P$ is the transition matrix from the basis used for $B$ to the basis used for $A$.

My argument for the claim  $[T]_{B} = [T]_{B'} \implies B = B'$ is as follows – can anyone help me spot the flaw? I feel I am misunderstanding the theorem.

Let $B = \{v_{1}, v_{2}, \ldots, v_{n}\}$ and $B'=\{w_{1}, w_{2}, \ldots, w_{n}\}$ be two bases for $\mathbb{R}^{n}$, and suppose that $[T]_{B} = [T]_{B'}$ for some linear operator $T$. Then 
  $$[T]_{B} = [T]_{B'} = I^{-1}[T]_{B'}I,$$ 
  so it follows from the theorem that $I$ is the transition matrix from $B$ to $B'$. Thus
  $$[v_{1}]_{B'} = I[v_{1}]_{B} = [v_{1}]_B = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, $$
  and so $v_{1} = w_{1}$. Similarly $v_{i} = w_{i}$ for each $i \in \{1, 2, \ldots, n\}$, so $B = B'$.


Edit: The full statement of the theorem in the book is as follows:

Theorem 8.5.3 If $V$ is a finite-dimensional vector space, then two matrices $A$ and $B$ represent the same linear operator (but possibly with respect to different bases) if and only if they are similar. Moreover, if $B = P^{-1}AP$, then $P$ is the transition matrix from the basis used for $B$ to the basis used for $A$.

 A: The theorem you've cited can be reformulated as follows. Let $P$ be the transition matrix between the basis $\beta'$ used for $B$ and the basis $\beta$ used for $A$. Let $T$ be a linear transformation, and let $[T]_\beta$ be represented by matrix $A$. Then the matrix $P^{-1} A P$ represents $[T]_{\beta'}$.
Note however, $P^{-1} A P = A$ (in your case, $P^{-1} B P = B'$) does not imply that $P = I$. Indeed, for every basis $\beta'$ with $[T]_{\beta'} = [T]_\beta$ this is true for the matrix $P$ whose column vectors are the vectors of $\beta$. What your proof shows is that if the transition matrix is $I$ (namely $\beta = \beta'$), then $[T]_\beta = [T]_{\beta'}$ (and this is of course trivial).
So to point out the specific incorrect statement, it would have to be "so it follows from the theorem that $I$ is the transition matrix from $B$ to $B'$." This is incorrect. Instead, the only statement you can make is "$I$ is one of the possible transition matrices from $B$ to $B'$." Hope this helps!
