0
$\begingroup$

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?

  • 0
    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 $T^{-1}$.2012-07-03
  • 0
    @ArturoMagidin So, $w_k = \sum^n_{k=1} (T^{-1})^j_k v_j$?2012-07-03
  • 0
    Yes; just switch the roles of $v_i$ and $w_j$ when using $T^{-1}$.2012-07-03
  • 0
    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.2012-07-03
  • 0
    @ArturoMagidin Thanks for the tips but I just don't see it; it seems backward to me.2012-07-03
  • 0
    @AFX: The matrix that transitions from basis $v$ to basis $w$ has, in the columns, the expressions for the vectors of $v$ in terms of the vectors of $w$. Your $T$ transitions from $v$ to $w$, so the columns of $T$ tell you how to write the vectors $v_1,\ldots,v_n$ in terms of the vectors $w_1,\ldots,w_n$. To see this, look at your equation $[T]^{v}_{w}[x]_v = [x]_w$. What is $[v_i]_v$? A vector that has a $1$ in the $i$th component as $0$s elsewhere. So what is $[T]^v_w[v_i]_v$? On the one hand, it is the $i$th column of $[T]^v_w$; on the other, it is $[v_i]_w$ ($v_i$ in terms of the $w_j$).2012-07-03

1 Answers 1