Matrix of Linear Transformation

As we have seen that any finite-dimensional vector space $V$ with an ordered basis is isomorphic to $\newcommand{R}{\mathbb{R}}\R^n$. Now suppose we have two vector spaces $V$ and $W$ over the same field $\R$, and a linear transformation $T$ between them. If we fix a basis $\alpha$ for $V$ and $\beta$ for $W$, then we have following situation

where the function $f$ is given by the compositions of these functions as follows \begin{align} f=\rho_\beta\circ T\circ \rho_{\alpha}^{-1} \end{align} So, for any linear transformation $T: V\to W$, we can find a unique linear transformation $f:\R^n\to \R^n$.

One important property of the vector spaces like $\R^n$ is that any linear transformation between such space is given by a matrix as the following theorem

Theorem: For any field $F$, any linear transformation $T:F^n \to F^m$ is given by a matrix, $A$, of size $m\times n$, such that, $T(\mathbf{x}) = A\mathbf{x}$.

##Matrix of a Linear Transformation

Definition: Let $V$ and $W$ be two vector spaces over a field $F$ with ordered basis $\alpha$ and $\beta$ respectively. Suppose $T: V\to W$ is a linear transformation, then the matrix associated with the linear transformation $f$ is called the matrix of $T$ with respect to $\alpha$ and $\beta$, and it is denoted by $\left[\, T \,\right ]_{\alpha}^{\beta}$.

The following theorem provides a method to calculate the matrix of a linear transformation with given bases.

Theorem: Let $T: V\to W$ be a linear transformation from a vector space $v$ to $w$ over field $F.$ Let $\alpha=(v_1, \dots, v_n)$ and $\beta=(w_1,\dots,w_n)$ be the ordered basis of $V$ and $W$ respectively. Then the matrix= of linear transformation $T$ w.r.to $\alpha$ and $\beta$ is given by.

Proof For any $v\in V$, we have

\begin{align} v&=x_1v_1+\dots+x_nv_n, \quad x_{i}\in F
\end{align}

Now applying $T$ on both sides, we get

\begin{align} T(v)&=T(x_{1}v_{1}+\dots+x_nv_n)
T(v)&=x_1T(v_1)+\dots+x_{n}T(v_n)
\end{align}

Since, $T(v_i)\in W$, we can find scalar $a_{ij}$, such that,

\begin{align} T(v_1)&=a_{11}w_1+a_{22}w_2+\dots+a_{m1}w_n
&\vdots
T(v_n)&=a_{1n}w_1+a_{2n}w_n+\dots+a_{mn} w_{m} \end{align}

Now assume,

\begin{align} (a_{11},\dots, a_{m1})&=[T(v_1)]_{\beta}
(a_{22},\dots, a_{m2})&=[T(v_2)]_{\beta}
&\vdots
(a_{m1},\dots, a_{mn})&=[T(v_n)]_{\beta} \end{align}

\begin{align} T(v)&=\sum_{i=1}^{n} X_{i}T(v_{i})
T(v_i)&=\sum_{v=1}^{n}a_{ji}w_{i} \end{align}

\begin{align} T(V)=\sum_{i=1}^{n}X_{i}T(V_{i})&=\sum_{i=1}^{n}x_i\left(\sum_{j=1}^{n}a_{ji}w_{j}\right)
&=\sum_{i=1}^{n}\sum_{j=1}^{m} x_i a_{ji}w_{j}
&=\sum_{j=1}^{m}\left(\sum_{i=1}^{n}x_{i}a_{ji}\right)w_{j}
&=\sum_{j=1}^{m}\left(\sum_{i=1}^{n}a_{ji}x_{i}\right)w_{j} \end{align}

\begin{align} \left[T(v)\right]_{\beta}&=\left(\sum_{i=1}^{n}a_{1i} X_{i}, \sum_{i=1}^{n}a_{2j}X_i, \dots, \sum_{i=1}^{n}a_{mi}X_{i}\right)
\left[T(v)\right]_{\beta}&=\left[\begin{array}{ccccc} \sum_{i=1}^{n}a_{1i} X_{i} &=a_{11}X_1&+a_{12}X_2&+\dots&+a_{1n}X_{n}
\sum_{i=1}^{n}a_{2i} X_{i} &=a_{21}X_1&+a_{22}X_2&+\dots&+a_{2n}X_{n}
\vdots & \vdots &\vdots & \vdots
\sum_{i=1}^{n}a_{1i} X_{i} &=a_{11}X_1&+a_{12}X_2&+\dots&+a_{1n}X_{n}
\end{array}\right]
&=\left[\begin{array}{cccc} a_11&\dots&a_{1n}
a_21& \dots& a_{2n}\ \vdots &\ddots & \vdots
a_{m1}&\dots & a_{mn} \end{array}\right]\left[\begin{array}{c} X_1
X_2
\vdots
X_{n} \end{array}\right]
&=\big [\left[T(v_1)_{\beta}\right] \dots \left[T(v_{n})_{\beta}\right] \big ]\left[V\right]_{\alpha} \end{align}

After watching this you can try question 1 from assignment 1.