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Chapter 2: Q2.7-10Q (page 93)

Consider the following geometric 2D transformations: D, a dilation (in which x-coordinates and y-coordinates are scaled by the same factor); R, a rotation; and T a translation. Does D commute with R? That is, is \(D\left( {R\left( {\bf{x}} \right)} \right) = R\left( {D\left( {\bf{x}} \right)} \right)\)for all \({\bf{x}}\) in \({\mathbb{R}^{\bf{2}}}\)? Does D commute with T? Does R commute with T?

Short Answer

Expert verified

D commutes with R, and matrices D and R do not commute with T.

Step by step solution

01

Write the transformation matrix for dilation, rotation, and translation

Let the transformation matrices in homogenous coordinates for dilation, rotation and translation be

\(D = \left[ {\begin{array}{*{20}{c}}s&0&0\\0&s&0\\0&0&1\end{array}} \right]\), \(R = \left[ {\begin{array}{*{20}{c}}{\cos \phi }&{ - \sin \phi }&0\\{\sin \phi }&{\cos \phi }&0\\0&0&1\end{array}} \right]\), and \(T = \left[ {\begin{array}{*{20}{c}}1&0&h\\0&1&k\\0&0&1\end{array}} \right]\).

02

Compute the product of transformation matrices

Calculate the value of \(DR\).

\(DR = \left[ {\begin{array}{*{20}{c}}{s\cos \phi }&{ - s\sin \phi }&0\\{s\sin \phi }&{s\cos \phi }&0\\0&0&1\end{array}} \right]\)

Calculate the value of \(RD\).

\[RD = \left[ {\begin{array}{*{20}{c}}{s\cos \phi }&{ - s\sin \phi }&0\\{s\sin \phi }&{s\cos \phi }&0\\0&0&1\end{array}} \right]\]

Calculate the value of \(DT\).

\(DT = \left[ {\begin{array}{*{20}{c}}s&0&{sh}\\0&s&{sk}\\0&0&1\end{array}} \right]\)

Calculate the value of \(TD\).

\(TD = \left[ {\begin{array}{*{20}{c}}s&0&h\\0&s&k\\0&0&1\end{array}} \right]\)

Calculate the value of \(RT\).

\(RT = \left[ {\begin{array}{*{20}{c}}{\cos \phi }&{ - \sin \phi }&{h\cos \phi - k\sin \phi }\\{\sin \phi }&{\cos \phi }&{h\sin \phi + k\cos \phi }\\0&0&1\end{array}} \right]\)

Calculate the value of \(TR\).

\(TR = \left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&h\\{\sin \varphi }&{\cos \varphi }&k\\0&0&1\end{array}} \right]\)

03

Check the commutation of transformation matrices

For the transformation matrices, \(DR = RD\).

So, D commutes with R.

\(DT \ne TD\)

So, D does not commute with T.

And

\(RT \ne TR\)

So, R does not commute with T.

Hence, D commutes with R, and matrices D and R do not commute with T.

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}3&{ - 6}\\{ - 1}&2\end{aligned}} \right)\). Construct a \({\bf{2}} \times {\bf{2}}\) matrix Bsuch that ABis the zero matrix. Use two different nonzero columns for B.

Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

8. \[\left[ {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\{\bf{0}}&{\bf{0}}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&I\end{array}} \right]\]

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).
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