Matrix of Linear Transformation¶

As we have seen that any finite-dimensional vector space $V$ with an ordered basis is isomorphic to $R^{n}$ . 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 $α$ for $V$ and $β$ for $W$ , then we have following situation

where the function $f$ is given by the compositions of these functions as follows $\begin{array}{r} f = ρ_{β} \circ T \circ ρ_{α}^{- 1} \end{array}$ 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, $T (x) = A x$ .

Matrix of a Linear Transformation¶

Definition: Let $V$ and $W$ be two vector spaces over a field $F$ with ordered basis $α$ and $β$ 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 $α$ and $β$ , and it is denoted by ${[T]}_{α}^{β}$ .

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 $α = (v_{1}, \dots, v_{n})$ and $β = (w_{1}, \dots, w_{n})$ be the ordered basis of $V$ and $W$ respectively. Then the matrix= of linear transformation $T$ w.r.to $α$ and $β$ is given by.

[T]_{α}^{β} = {[\begin{matrix} ⋮ & ⋮ & ⋮ \\ [T (v_{1})]_{β} & [T (v_{2})]_{β} & \dots & [T (v_{n})]_{β} \\ ⋮ & ⋮ & ⋮ \end{matrix}]}_{m \times n}

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

\begin{aligned} v & = x_{1} v_{1} + \dots + x_{n} v_{n}, x_{i} \in F \end{aligned}

Now applying $T$ on both sides, we get

\begin{aligned} T (v) & = T (x_{1} v_{1} + \dots + x_{n} v_{n}) \\ T (v) & = x_{1} T (v_{1}) + \dots + x_{n} T (v_{n}) \end{aligned}

Since, $T (v_{i}) \in W$ , we can find scalar $a_{i j}$ , such that,

\begin{aligned} T (v_{1}) & = a_{11} w_{1} + a_{22} w_{2} + \dots + a_{m 1} w_{n} \\ ⋮ \\ T (v_{n}) & = a_{1 n} w_{1} + a_{2 n} w_{n} + \dots + a_{m n} w_{m} \end{aligned}

Now assume,

\begin{aligned} (a_{11}, \dots, a_{m 1}) & = [T (v_{1})]_{β} \\ (a_{22}, \dots, a_{m 2}) & = [T (v_{2})]_{β} \\ ⋮ \\ (a_{m 1}, \dots, a_{m n}) & = [T (v_{n})]_{β} \end{aligned}

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

\begin{aligned} T (V) = \sum_{i = 1}^{n} X_{i} T (V_{i}) & = \sum_{i = 1}^{n} x_{i} (\sum_{j = 1}^{n} a_{j i} w_{j}) \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{m} x_{i} a_{j i} w_{j} \\ = \sum_{j = 1}^{m} (\sum_{i = 1}^{n} x_{i} a_{j i}) w_{j} \\ = \sum_{j = 1}^{m} (\sum_{i = 1}^{n} a_{j i} x_{i}) w_{j} \end{aligned}

\begin{aligned} {[T (v)]}_{β} & = (\sum_{i = 1}^{n} a_{1 i} X_{i}, \sum_{i = 1}^{n} a_{2 j} X_{i}, \dots, \sum_{i = 1}^{n} a_{m i} X_{i}) \\ {[T (v)]}_{β} & = [\begin{array}{ccccc} \sum_{i = 1}^{n} a_{1 i} X_{i} & = a_{11} X_{1} & + a_{12} X_{2} & + \dots & + a_{1 n} X_{n} \\ \sum_{i = 1}^{n} a_{2 i} X_{i} & = a_{21} X_{1} & + a_{22} X_{2} & + \dots & + a_{2 n} X_{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ \sum_{i = 1}^{n} a_{1 i} X_{i} & = a_{11} X_{1} & + a_{12} X_{2} & + \dots & + a_{1 n} X_{n} \end{array}] \\ = [\begin{array}{cccc} a_{1} 1 & \dots & a_{1 n} \\ a_{2} 1 & \dots & a_{2 n} \\ ⋮ & ⋱ & ⋮ \\ a_{m 1} & \dots & a_{m n} \end{array}] [\begin{array}{c} X_{1} \\ X_{2} \\ ⋮ \\ X_{n} \end{array}] \\ = [[T (v_{1})_{β}] \dots [T (v_{n})_{β}]] {[V]}_{α} \end{aligned}

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

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