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Composition and Inverse

Composition of Transformations

Suppose we have linear transformations T:RkRnT:\mathbb{R}^k\to\mathbb{R}^n and S:RnRmS:\mathbb{R}^n\to\mathbb{R}^m.
The composition of SS and TT is another linear transformation denoted ST:RkRmS \circ T: \reals^k \to \reals^m. It is defined as:
(ST)(x)=S(T(x))\boxed{\quad \left(S\circ T\right)(\vec{x})=S(T(\vec{x})) \quad}
Wize Tip
To determine the right order in which to apply transformations, read the composition from right to left!
E.g. (UV)(x)(U \circ V )(\vec x) means find V(x)\colorOne{V(\vec x)} first, then use that result to find U(V(x))U(\, \colorOne{V(\vec x)}\, ).

If AA is the matrix inducing TT, and BB is the matrix inducing SS, then BABA is the matrix inducing STS\circ T:
(ST)(x)=S(T(x))=B(Ax)=(BA)x\begin{aligned} \left(S\circ T\right)(\vec{x}) &=S(T(\vec{x})) \\[0.3em] &=B(A\vec{x}) \\[0.3em] &= (BA)\vec{x} \end{aligned}
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Inverse of a Linear Transformation

Suppose T:RnRnT:\mathbb{R}^n\to\mathbb{R}^n and S:RnRnS:\mathbb{R}^n\to\mathbb{R}^n are linear transformations.
We say that SS and TT are inverses of one another if, for every xRn\vec{x}\in \reals^n:
(ST)(x)=x(S\circ T)(\vec{x})=\vec{x} and (T S)(x)=x(T\circ\ S)(\vec{x})=\vec{x}

Wize Tip
Geometrically, the inverse transformation "undoes" the original transformation.
E.g. The inverse of a rotation by 20°20 \degree is a rotation by 20°-20 \degree.

Matrix Form of the Inverse Transformation

If AA is the matrix inducing TT, then TT has an inverse transformation if and only if AA is invertible.
The inverse is unique and denoted by T1:RnRnT^{-1}:\mathbb{R}^n\to\mathbb{R}^n , and is induced by the matrix A1A^{-1}.
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Example: Composition of Linear Transformations

Suppose TT and SS are linear transformations induced by the matrices A=[2024], B=[1021]A= \begin{bmatrix} -2&0\\ 2&4 \end{bmatrix}, \ B= \begin{bmatrix} 1&0\\ -2&1 \end{bmatrix}, respectively.

Part 1)

If x=[15]\vec{x}= \begin{bmatrix} 1\\ 5 \end{bmatrix}, find (ST)(x)(S \circ T)(\vec{x}).
(ST)(x)=BAx=[1021][2024][15]=[2064][15]=[226]\begin{aligned} (S \circ T)(\vec x) &= BA\vec x\\[0.3em] &= \begin{bmatrix} 1&0\\ -2&1 \end{bmatrix} \begin{bmatrix} -2&0\\ 2&4 \end{bmatrix} \begin{bmatrix} 1\\ 5 \end{bmatrix}\\[1.5em] &= \colorTwo{ \begin{bmatrix} -2&0\\ 6&4 \end{bmatrix}} \begin{bmatrix} 1\\ 5 \end{bmatrix}\\[1.5em] &= \boxed{ \begin{bmatrix} -2\\ 26 \end{bmatrix}} \end{aligned}
Note that AB=[2064]AB= \colorTwo{ \begin{bmatrix} -2&0\\ 6&4 \end{bmatrix}} is the matrix that induces STS \circ T.

Part 2)

Find the matrix inducing TS1T \circ S^{-1}.
Since SS is induced by the matrix BB, S1S^{-1} must be induced by B1B^{-1}.
B=[1021]    B1 = [10(2)1] = [1021]B= \begin{bmatrix} \colorOne1 & 0\\ -2 & \colorTwo1 \end{bmatrix} \quad\implies\quad B^{-1} \ =\ \begin{bmatrix} \colorTwo1& \colorFour{-}0\\ \colorFour{-}(-2) & \colorOne1 \end{bmatrix} \ =\ \begin{bmatrix} 1 & 0\\ 2 &1 \end{bmatrix}
We know that for any vector xR2\vec x \in \reals^2:
(TS1)(x)=T(S1(x))=T(B1x)=AB1x(T \circ S^{-1})(\vec x) = T(\, S^{-1}(\vec x)\, ) = T(B^{-1}\vec x) = AB^{-1}\vec x
Therefore, the matrix inducing TS1T \circ S^{-1} is given by AB1AB^{-1}:
AB1=[2024][1021]=[20104]AB^{-1} = \begin{bmatrix} -2&0\\ 2&4 \end{bmatrix} \begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix} = \boxed{ \begin{bmatrix} -2&0\\ 10&4 \end{bmatrix} }

Practice: Composition of Linear Transformations

Let T:R2R3T:\mathbb{R}^2\to\mathbb{R}^3 be a linear transformation defined by T[xy]=[x+y3yx+2y]T\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x+y\\3y\\x+2y\end{bmatrix}.
Let S:R2R2S:\reals^2 \to \reals^2 be the linear transformation induced by the matrix B=[1101]B= \left[ \begin{array}{rrr} 1 & 1 \\ 0 & 1 \\ \end{array} \right].
Find the vector u=[uv]\vec u = \begin{bmatrix} u\\ v \end{bmatrix} such that (TS1)(u)=[231](T\circ S^{-1})(\vec u) = \begin{bmatrix} 2\\ -3\\ 1 \end{bmatrix}.
Extra Practice