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Let\(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\) and \(D = \left\{ {{{\bf{d}}_{\bf{1}}},{{\bf{d}}_{\bf{2}}}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\)

Find the matrix for \(T\) relative to \(B\), and\(D\).

Short Answer

Expert verified

The matrix for \(T\) relative to \(B\) and \(D\) is \(\left( {\begin{aligned}3&{}&{ - 1}&{}&0\\{ - 5}&{}&6&{}&4\end{aligned}} \right)\).

Step by step solution

01

Matrix for a linear transformation 

A matrix associated with a linear transformation \(T\) for \(V\) and \(W\)is given by \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(V\) and \(W\) are \(n\) and \(m\)-dimensional subspaces respectively, and \(B\), and \(C\) are the bases for \(V\), and \(W\).

02

Find the matrix for a linear transformation 

It is given that \(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\) and \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\).

Write them in the form of vectors as shown below:

\({\left( {T\left( {{{\bf{b}}_1}} \right)} \right)_D} = \left( {\begin{aligned}3\\{ - 5}\end{aligned}} \right)\)

\({\left( {T\left( {{{\bf{b}}_2}} \right)} \right)_D} = \left( {\begin{aligned}{ - 1}\\6\end{aligned}} \right)\)

\({\left( {T\left( {{{\bf{b}}_3}} \right)} \right)_D} = \left( {\begin{aligned}0\\4\end{aligned}} \right)\)

Form a matrix\(T\) for the obtained vectors by using the formula \({\left( {T\left( {\bf{x}} \right)} \right)_C} = \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_C}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_C}}& \cdots &{{{\left( {T\left( {{{\bf{b}}_n}} \right)} \right)}_C}}\end{aligned}} \right)\), where \(n = 3\).

\(\begin{aligned}{c}{\left( {T\left( {\bf{x}} \right)} \right)_D} &= \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{b}}_1}} \right)} \right)}_D}}&{{{\left( {T\left( {{{\bf{b}}_2}} \right)} \right)}_D}}&{{{\left( {T\left( {{{\bf{b}}_3}} \right)} \right)}_D}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}3&{}&{ - 1}&{}&0\\{ - 5}&{}&6&{}&4\end{aligned}} \right)\end{aligned}\)

So, the required matrix is \(\left( {\begin{aligned}3&{}&{ - 1}&{}&0\\{ - 5}&{}&6&{}&4\end{aligned}} \right)\).

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Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left[ {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right]\).

22. Let \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + {a_{\bf{2}}}{t^{\bf{2}}} + {t^{\bf{3}}}\), and let \(\lambda \) be a zero of \(p\).

  1. Write the companion matrix for \(p\).
  2. Explain why \({\lambda ^{\bf{3}}} = - {a_{\bf{0}}} - {a_{\bf{1}}}\lambda - {a_{\bf{2}}}{\lambda ^{\bf{2}}}\), and show that \(\left( {{\bf{1}},\lambda ,{\lambda ^2}} \right)\) is an eigenvector of the companion matrix for \(p\).

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

16. \(\left( {\begin{array}{*{20}{c}}{\bf{0}}&{{\bf{ - 4}}}&{{\bf{ - 6}}}\\{{\bf{ - 1}}}&{\bf{0}}&{{\bf{ - 3}}}\\{\bf{1}}&{\bf{2}}&{\bf{5}}\end{array}} \right)\)

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