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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 \(T\) \(\alpha\) 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.

Last modified on: 2023-01-11 01:50:45