Question

How to find the transformation matrix of a basis?

Answer 1

Let $ V$ and $ W$ be finite dimensional vector spaces over the set $ {\mathbb{F}}$ with respective dimensions $ m$ and $ n.$ Also, let $ T: V {\longrightarrow}W$ be a linear transformation. Suppose $ {\cal B}_1=( {\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n )$ is an ORDERED BASIS of $ V.$ In the last section, we saw that a linear transformation is determined by its image on a basis of the domain space. We therefore look at the images of the vectors $ {\mathbf v}_j \in {\cal B}_1$ for $ 1 \leq j \leq n.$

Now for each $ j, \; 1 \leq j \leq n,$ the vectors $ T({\mathbf v}_j) \in W.$ We now express these vectors in terms of an ordered basis $ {\cal B}_2= ( {\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m)$ of $ W.$ So, for each $ j, \; 1 \leq j \leq n,$ there exist unique scalars $ a_{1j}, a_{2j}, \ldots, a_{mj} \in {\mathbb{F}}$ such that
$\displaystyle T({\mathbf v}_1)$ $\displaystyle =$ $\displaystyle a_{11} {\mathbf w}_1 + a_{21} {\mathbf w}_2 +
\cdots + a_{m1} {\mathbf w}_m$
$\displaystyle T({\mathbf v}_2)$ $\displaystyle =$ $\displaystyle a_{12} {\mathbf w}_1 + a_{22} {\mathbf w}_2 + \cdots
+ a_{m2} {\mathbf w}_m$
$\displaystyle \vdots$
$\displaystyle T({\mathbf v}_n)$ $\displaystyle =$ $\displaystyle a_{1n} {\mathbf w}_1 + a_{2n} {\mathbf w}_2 +
\cdots + a_{mn} {\mathbf w}_m.$

Or in short, $ T({\mathbf v}_j) =
\sum\limits_{i=1}^m a_{ij} {\mathbf w}_i$ for $ 1 \leq j \leq n.$ In other words, for each $ j, \; 1 \leq j \leq n,$ the coordinates of $ T({\mathbf v}_j) $ with respect to the ordered basis $ {\cal B}_2$ is the column vector $ [a_{1j}, a_{2j},
\ldots, a_{mj}]^t.$ Equivalently,
$\displaystyle [T({\mathbf v}_j)]_{{\cal B}_2} = \begin{bmatrix}
a_{1j} \\ a_{2j} \\ \vdots \\ a_{mj} \end{bmatrix}.$
Let $ [{\mathbf x}]_{{\cal B}_1} = [x_1, x_2, \ldots, x_n]^t$ be the coordinates of a vector $ {\mathbf x}\in V.$ Then
$\displaystyle T({\mathbf x})$ $\displaystyle =$ $\displaystyle T ( \sum_{j=1}^n x_j {\mathbf v}_j) = \sum_{j=1}^n x_j T({\mathbf v}_j)$
$\displaystyle =$ $\displaystyle \sum_{j=1}^n x_j
( \sum_{i=1}^m a_{ij} {\mathbf w}_i)$
$\displaystyle =$ $\displaystyle \sum_{i=1}^m (
\sum\limits_{j=1}^n a_{ij} x_j) {\mathbf w}_i.$

Define a matrix $ A$ by $ A = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & ...
... \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}.$ Then the coordinates of the vector $ T({\mathbf x})$ with respect to the ordered basis $ {\cal B}_2$ is
$\displaystyle [T({\mathbf x})]_{{\cal B}_2}$ $\displaystyle =$ $\displaystyle \begin{bmatrix}\sum_{j=1}^n a_{1j} x_j \\
\sum_{j=1}^n a_{2j} x_...
... & a_{mn} \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$
$\displaystyle =$ $\displaystyle A \; [{\mathbf x}]_{{\cal B}_1}.$

The matrix $ A$ is called the matrix of the linear transformation $ T$ with respect to the ordered bases $ {\cal B}_1$ and $ {\cal B}_2,$ and is denoted by $ T[{\cal B}_1, {\cal B}_2].$