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Chapter 2: Q2.7-12Q (page 93)

A rotation in \({\mathbb{R}^{\bf{2}}}\) usually requires four multiplications. Compute the product below, and show that the matrix for a rotation can be factored into three shear transformations (each of which requires only one multiplication).

\(\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{ - tan\varphi /2}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\end{array}} \right]\)\[\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}\\{sin\varphi }&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\end{array}} \right]\] \(\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{ - tan\varphi /2}&{\bf{0}}\\{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\end{array}} \right]\)

Short Answer

Expert verified

The composition is \[\left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\]. For a rotation in \({\mathbb{R}^2}\), the composition is the transformation matrix in coordinates.

Step by step solution

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01

State the row-column rule

If the sum of the products of matching entries from row\(i\)of matrix A and column\(j\)of matrix B equals the item in row\(i\)and column\(j\)of AB, then it can be said that product AB is defined.

The product is \({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\).a

02

Obtain the product of the matrices

Consider the matrices\(\left[ {\begin{array}{*{20}{c}}1&{ - \tan \varphi /2}&0\\0&1&0\\0&0&1\end{array}} \right]\)and\[\left[ {\begin{array}{*{20}{c}}1&0&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\].

Obtain the product of the first two matrices, as shown below:

\(\begin{array}{c} \Rightarrow \left[ {\begin{array}{*{20}{c}}1&{ - \tan \varphi /2}&0\\0&1&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1\left( 1 \right) - \tan \varphi /2\left( {\sin \varphi } \right) + 0\left( 0 \right)}&{1\left( 0 \right) - \tan \varphi /2\left( 1 \right) + 0\left( 0 \right)}&{1\left( 0 \right) - \tan \varphi /2\left( 0 \right) + 0\left( 1 \right)}\\{0\left( 1 \right) + 1\left( {\sin \varphi } \right) + 0\left( 0 \right)}&{0\left( 0 \right) - 1\left( 1 \right) + 0\left( 0 \right)}&{0\left( 0 \right) + 1\left( 0 \right) + 0\left( 1 \right)}\\{0\left( 1 \right) + 0\left( {\sin \varphi } \right) + 1\left( 0 \right)}&{0\left( 0 \right) + 1\left( 1 \right) + 0\left( 0 \right)}&{0\left( 0 \right) + 0\left( 0 \right) + 0\left( 1 \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{ - \tan \varphi /2}&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\end{array}\)

Thus,\(\left[ {\begin{array}{*{20}{c}}1&{ - \tan \varphi /2}&0\\0&1&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{ - \tan \varphi /2}&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\).

Now, multiply the matrix \(\left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{ - \tan \varphi /2}&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\) by \(\left[ {\begin{array}{*{20}{c}}1&{ - \tan \varphi /2}&0\\0&1&0\\0&0&1\end{array}} \right]\), as shown below:

\[\begin{array}{c} \Rightarrow \left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{ - \tan \varphi /2}&0\\{\sin \varphi }&1&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - \tan \varphi /2}&0\\0&1&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{\left( {1 - \sin \varphi \tan \varphi /2} \right)\left( 1 \right) - \tan \varphi /2\left( 0 \right)}&{\left( {1 - \sin \varphi \tan \varphi /2} \right)\left( { - \tan \varphi /2} \right) - \tan \varphi /2\left( 1 \right)}&0\\{\sin \varphi \left( 1 \right) + 1\left( 0 \right) + 0\left( 0 \right)}&{\sin \varphi \left( { - \tan \varphi /2} \right) + 1\left( 1 \right) + 0\left( 0 \right)}&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{\sin \varphi {{\tan }^2}\varphi /2 - 2\tan \varphi /2}&0\\{\sin \varphi }&{ - \sin \varphi \tan \varphi /2 + 1}&0\\0&0&1\end{array}} \right]\end{array}\]

Thus, the product is \[\left[ {\begin{array}{*{20}{c}}{1 - \sin \varphi \tan \varphi /2}&{\sin \varphi {{\tan }^2}\varphi /2 - 2\tan \varphi /2}&0\\{\sin \varphi }&{1 - \sin \varphi \tan \varphi /2}&0\\0&0&1\end{array}} \right]\].

03

Use the trigonometric identities

Simplify the trigonometric entry\[1 - \sin \varphi \tan \varphi /2\]using a trigonometric identity, as shown below:

\[\begin{array}{c}1 - \sin \varphi \tan \varphi /2 = 1 - \frac{{2{{\tan }^2}\varphi /2}}{{1 + {{\tan }^2}\varphi /2}}\\ = \frac{{1 + {{\tan }^2}\varphi /2 - 2{{\tan }^2}\varphi /2}}{{1 + {{\tan }^2}\varphi /2}}\\ = \frac{{1 - {{\tan }^2}\varphi /2}}{{1 + {{\tan }^2}\varphi /2}}\\ = \cos \varphi \end{array}\]

Simplify \[\sin \varphi {\tan ^2}\varphi /2 - 2\tan \varphi /2\].

\[\begin{array}{c}\sin \varphi {\tan ^2}\varphi /2 - 2\tan \varphi /2 = \tan \varphi /2\left( {\sin \varphi \tan \varphi /2 - 2} \right)\\ = \tan \varphi /2\left( {\frac{{2{{\tan }^2}\varphi /2}}{{1 + {{\tan }^2}\varphi /2}} - 2} \right)\\ = \tan \varphi /2\left( {\frac{{2{{\tan }^2}\varphi /2 - 2 - 2{{\tan }^2}\varphi /2}}{{1 + {{\tan }^2}\varphi /2}}} \right)\\ = - \frac{{\tan \varphi /2}}{{1 + {{\tan }^2}\varphi /2}}\\ = - \sin \varphi \end{array}\]

The product becomes

\[\left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right]\].

Therefore, for a rotation in \({\mathbb{R}^2}\), the composition is the transformation matrixin coordinates.

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Most popular questions from this chapter

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

In exercises 11 and 12, mark each statement True or False. Justify each answer.

a. The definition of the matrix-vector product \(A{\bf{x}}\) is a special case of block multiplication.

b. If \({A_{\bf{1}}}\), \({A_{\bf{2}}}\), \({B_{\bf{1}}}\), and \({B_{\bf{2}}}\) are \(n \times n\) matrices, \[A = \left[ {\begin{array}{*{20}{c}}{{A_{\bf{1}}}}\\{{A_{\bf{2}}}}\end{array}} \right]\] and \(B = \left[ {\begin{array}{*{20}{c}}{{B_{\bf{1}}}}&{{B_{\bf{2}}}}\end{array}} \right]\), then the product \(BA\) is defined, but \(AB\) is not.

Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\)as\(n \times 1\)matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is a \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

28. If u and v are in \({\mathbb{R}^n}\), how are \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \) related? How are \({{\mathop{\rm uv}\nolimits} ^T}\) and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\) related?

Use partitioned matrices to prove by induction that for \(n = 2,3,...\), the \(n \times n\) matrices \(A\) shown below is invertible and \(B\) is its inverse.

\[A = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\1&1&0&{}&0\\1&1&1&{}&0\\ \vdots &{}&{}& \ddots &{}\\1&1&1& \ldots &1\end{array}} \right]\]

\[B = \left[ {\begin{array}{*{20}{c}}1&0&0& \cdots &0\\{ - 1}&1&0&{}&0\\0&{ - 1}&1&{}&0\\ \vdots &{}& \ddots & \ddots &{}\\0&{}& \ldots &{ - 1}&1\end{array}} \right]\]

For the induction step, assume A and Bare \(\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrices, and partition Aand B in a form similar to that displayed in Exercises 23.
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