# 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.

$[\,T\,]^{\beta}_{\alpha}=\begin{bmatrix} \vdots &\vdots& &\vdots \\ [\,T(v_1)\,]_{\beta} & [\,T(v_2)\,]_{\beta} & \cdots & [\,T(v_n)\,]_{\beta} \\ \vdots &\vdots& &\vdots \end{bmatrix}_{m \times n}$

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.