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I wish to check my understanding on part of the proof of Proposition 5.3 in Lee's Introduction to Smooth Manifold. It reads as follows: $\def\tE {\widetilde{E}}$

Let $(E_i)$ and $(\tE_i)$ be two ordered bases for a finite-dimensional vector space $V$ and let $B:V\to V$ be the linear map sending $E_j$ to $\tE_j$. This means that $$\tE_j = BE_j = \sum_iB_i^j E_i,$$ so $B$ is the transition matrix between the two bases.

To me it does not even make sense to write $B$ in matrix form before stating on which basis such matrix form is being written. What is basis independent is the transition matrix $A^j_i$ between the two bases. So, I think, what the author means is that, in the basis $(E_i)$, the map $B$ has matrix matrix form $A_i^j$. Right?

Sam
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  • It's tacitly assumed that $B$ is written with respect to $(E_i$). This is typical in situations when you're working with two bases rather than three or more. – CyclotomicField May 09 '24 at 13:09
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    Aside: Here's how I like to see it (warning: different notation conventions than Lee): I like to write a basis of an abstract vector space as a row matrix of vectors. In other words, I write a basis $(e_1, \dots, e_n)$ as $$E=\begin{bmatrix}e_1 &\cdots & e_n\end{bmatrix}.$$ Given bases $E,\widetilde{E}$, there is an invertible matrix $B$ such that $\widetilde{E}=EB.$ Any vector $v$ can be written as $v=Ea$, where $a$ is a column matrix. The change of basis formula for $v$ is now easily derived: $$v=\widetilde{E}\tilde{a}=EB\tilde{a}=Ea,$$where $a=B\tilde{a}$ or $\tilde{a}=B^{-1}a$. – Deane May 09 '24 at 14:43
  • Have a look at the definition of a linear map and its associated matrix in an introductory book like Linear Algebra Done Right by Sheldon Axler. Any linear map can be described by its action on the basis of a vector space. So given $B e_i = {B^j}_i{\tilde e}_j$ we have the complete definition of this linear map. ${B^j}_i$ is the corresponding matrix; in the same way that $x^i$ in the expansion $x=x^i e_i$ describes the coordinates of vector $x$ under basis $E$, ${B^j}_i$ describes the matrix associated with the transformation from basis $E$ to basis $\tilde E$. – Ted Black May 09 '24 at 17:08

2 Answers2

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This is Proposition 15.3, not 5.3.

You're right, saying that $B$ "is" the transition matrix is incorrect. What I should have said is that the matrix representation of $B$ with respect to $(E_i)$ is the transition matrix between the two bases. Thanks for pointing this out. I've added it to my online correction list.

Jack Lee
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Things are more clear if you get a bit notation-heavy. I'm going to use $F$ instead of $\widetilde{E}$ because it's easier to type.

If I have a linear operator $B: U \to V$, and a basis $(E_i)$ of $U$ and $(F_i )$ of $V$, then define the matrix $[^F B ^E]_{ij}$ by the equations $$ B E_j = \sum_i [^F B ^E]_{ij} F_i \quad \text{ for all } j.$$ Now you should be able to see that if $U = V$ and $BE_i = F_i$ (as in the statement), that

  1. $[^F B ^E]$ is the identity matrix.
  2. $[^E B ^E]$ is the matrix the author is talking about.
Joppy
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