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Let\(D = \left\{ {{{\bf{d}}_1},{{\bf{d}}_2}} \right\}\) and \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\)

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

Short Answer

Expert verified

The matrix for \(T\) relative to \(D\) and \(B\) is \(\left( {\begin{aligned}2&{}&{ - 4}\\{ - 3}&{}&5\end{aligned}} \right)\).

Step by step solution

01

The 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 \(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), and \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\).

Write them in the form of vectors as shown below:

\({\left( {T\left( {{{\bf{d}}_1}} \right)} \right)_B} = \left( {\begin{aligned}2\\{ - 3}\end{aligned}} \right)\)

\({\left( {T\left( {{{\bf{d}}_2}} \right)} \right)_B} = \left( {\begin{aligned}{ - 4}\\5\end{aligned}} \right)\)

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 \(n = 2\).

\(\begin{aligned}{\left( {T\left( {\bf{x}} \right)} \right)_B} &= \left( {\begin{aligned}{{{\left( {T\left( {{{\bf{d}}_1}} \right)} \right)}_B}}&{{{\left( {T\left( {{{\bf{d}}_2}} \right)} \right)}_B}}\end{aligned}} \right)\\ &= \left( {\begin{aligned}2&{}&{ - 4}\\{ - 3}&{}&5\end{aligned}} \right)\end{aligned}\)

So, the requiredmatrix is \(\left( {\begin{aligned}2&{}&{ - 4}\\{ - 3}&{}&5\end{aligned}} \right)\).

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Most popular questions from this chapter

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) an eigenvector of\(\left){\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\)? If so, find the eigenvalue.

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

  1. \(\left[ {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right]\)

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

Assume the mapping\(T:{{\rm P}_2} \to {{\rm P}_{\bf{2}}}\)defined by \(T\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right) = 3{a_0} + \left( {5{a_0} - 2{a_1}} \right)t + \left( {4{a_1} + {a_2}} \right){t^2}\) is linear. Find the matrix representation of\(T\) relative to the bases \(B = \left\{ {1,t,{t^2}} \right\}\).

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

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