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I have read on this site and from various sources trying to get an understanding of what the change of basis matrix and change of basis operator is. Below, I have given the definitions as I understand them and hope that someone can either confirm or correct my understanding of these things.

Part I

Let $v := (v_1, \dots, v_n)$ and $w := (w_1, \dots, w_n)$ be two bases of a vector space $V$. By a theorem, there exists a unique isomorphism $T:V \rightarrow V$ given by $T(v_i) = w_i, i=1, \dots n$. Let $[T]^v_w$ denote the matrix of $T$ with respect to $v$ and $w$. Then for any $x \in V$, by another theorem, $$ [T]^v_w[x]_v = [x]_w $$ where $[x]_b$ denotes the column matrix of $x$ with respect to a basis $b$. The matrix $[T]^v_w$ is called the change of basis matrix or the transition matrix from $v$ to $w$ and $T$ is called the transition automorphism or transition operator from $v$ to $w$.

Question: Is everything in the above acceptably defined?

Part II Looking at the above componentwise, for a vector $x = a^1v_1 + \dots +a^nv_n$ expressed in terms of the basis $v$, the $k$-th coordinate of $[x]_w$ is given by $$ ([x]_w)_k = \sum_{j = 1}^n T^k_ja^j $$ where $T^k_j$ denotes the entry of $T$ that lives in the $k$-th row and $j$-th column of $[T]^v_w$.

Question: How are the actual basis vectors related to one another with respect to the components of $T$? I think it should be $w_k = \sum_{k=1}^n T^j_k v_j$. How does one prove this rigorously? I believe it follows from the fact that the $T(v_k)$ corresponds to the $k^th$ column of $T$ which is represented by the scalars $T^j_k, k=1, \dots, n$. Is this right?

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Your last question: No; what you have is that $v_k = \sum T_k^jw_j$. To see this, note simply that $[v_k]_v$ has a $1$ in the $k$th entry and $0$s elsewhere; this will give you $[v_k]_w$. To express the $w_i$ in terms of the $v_j$'s, you need $T^{-1}$. –  Arturo Magidin Jul 3 '12 at 18:05
    
@ArturoMagidin So, $w_k = \sum^n_{k=1} (T^{-1})^j_k v_j$? –  AFX Jul 3 '12 at 18:08
    
Yes; just switch the roles of $v_i$ and $w_j$ when using $T^{-1}$. –  Arturo Magidin Jul 3 '12 at 18:08
    
Side note: in future, it's much easier if you split a long question into several questions posted over spaced time period. It works better & you'll get more (quantity & quality) feedback. –  user2468 Jul 3 '12 at 18:33
    
@ArturoMagidin Thanks for the tips but I just don't see it; it seems backward to me. –  AFX Jul 3 '12 at 18:59

1 Answer 1

The elements of $j$-th column of $T$, $t_{1j}, t_{2j}, \cdots, t_{nj}$, are the expansion coefficients of the $j$-th basis $v_j$ with respect to new basis $w_1,\cdots, w_n$, i.e. $v_j = \sum_{i=1}^nt_{i,j}w_i$. Likewise, the elements of $i$-th row of $T$, $t_{i1}, t_{i2}, \cdots, t_{in}$, are the expansion coefficients of the $i$-th basis $w_i$ with respect to old basis $v_1,\cdots, v_n$, i.e. $w_i = \sum_{j=1}^nt_{i,j}v_j$.

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