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Chapter 5: Q5.4-8E (page 267)

Let\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space\(V\). Find \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)\) when \(T\) isa linear transformation from \(V\) to \(V\) whose matrix relative to \(B\) is

\({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\)

Short Answer

Expert verified

The value of\({\left( {T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)} \right)_B}\)is \(\left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\), and \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right) = 24{{\bf{b}}_1} - 20{{\bf{b}}_2} + 11{{\bf{b}}_3}\).

Step by step solution

01

The vector form

Let \({\left( {\bf{x}} \right)_B} = 3{{\bf{b}}_1} - 4{{\bf{b}}_2}\), which can also be written as \({\left( {\bf{x}} \right)_B} = 3{{\bf{b}}_1} - 4{{\bf{b}}_2} + 0{{\bf{b}}_3}\).

Write \({\left( {\bf{x}} \right)_B}\) in the vector form.

\({\left( {\bf{x}} \right)_B} = \left( {\begin{aligned}3\\{ - 4}\\0\end{aligned}} \right)\)

02

Linear transformation forms \(V\) into \(V\) 

The \(B\)-matrix for \(T\) or a matrix for \(T\) relative to \(B\) for \(T:V \to V\) is given by,

\({\left( {T\left( {\bf{x}} \right)} \right)_B} = {\left( T \right)_B}{\left( {\bf{x}} \right)_B}\)

03

Find linear transformation form \(V\) into \(V\)

The given matrix is \({\left( T \right)_B} = \left( {\begin{aligned}0&{}&{ - 6}&{}&1\\0&{}&5&{}&{ - 1}\\1&{}&{ - 2}&{}&7\end{aligned}} \right)\).

Write\({\left( {T\left( {\bf{x}} \right)} \right)_B} = {\left( T \right)_B}{\left( {\bf{x}} \right)_B}\).

\(\begin{aligned}{\left( {T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)} \right)_B} &= {\left( T \right)_B}{\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right)_B}\\ &= \left( {\begin{aligned}0&{ - 6}&1\\0&5&{ - 1}\\1&{ - 2}&7\end{aligned}} \right)\left( {\begin{aligned}3\\{ - 4}\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{0\left( 3 \right) + \left( { - 6} \right)\left( { - 4} \right) + 1\left( 0 \right)}\\{0\left( 3 \right) + \left( 5 \right)\left( { - 4} \right) + \left( { - 1} \right)\left( 0 \right)}\\{1\left( 3 \right) + \left( { - 2} \right)\left( { - 4} \right) + 7\left( 0 \right)}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\end{aligned}\)

So, the required matrix is \(\left( {\begin{aligned}{24}\\{ - 20}\\{11}\end{aligned}} \right)\).

Thus, \(T\left( {3{{\bf{b}}_1} - 4{{\bf{b}}_2}} \right) = 24{{\bf{b}}_1} - 20{{\bf{b}}_2} + 11{{\bf{b}}_3}\).

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Most popular questions from this chapter

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

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 \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

23. Let \(p\) be the polynomial in Exercise \({\bf{22}}\), and suppose the equation \(p\left( t \right) = {\bf{0}}\) has distinct roots \({\lambda _{\bf{1}}},{\lambda _{\bf{2}}},{\lambda _{\bf{3}}}\). Let \(V\) be the Vandermonde matrix

\(V{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{1}}\\{{\lambda _{\bf{1}}}}&{{\lambda _{\bf{2}}}}&{{\lambda _{\bf{3}}}}\\{\lambda _{\bf{1}}^{\bf{2}}}&{\lambda _{\bf{2}}^{\bf{2}}}&{\lambda _{\bf{3}}^{\bf{2}}}\end{aligned}} \right)\)

(The transpose of \(V\) was considered in Supplementary Exercise \({\bf{11}}\) in Chapter \({\bf{2}}\).) Use Exercise \({\bf{22}}\) and a theorem from this chapter to deduce that \(V\) is invertible (but do not compute \({V^{{\bf{ - 1}}}}\)). Then explain why \({V^{{\bf{ - 1}}}}{C_p}V\) is a diagonal matrix.

Let\(\varepsilon = \left\{ {{{\bf{e}}_1},{{\bf{e}}_2},{{\bf{e}}_3}} \right\}\) be the standard basis for \({\mathbb{R}^3}\),\(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\) be a basis for a vector space \(V\) and\(T:{\mathbb{R}^3} \to V\) be a linear transformation with the property that

\(T\left( {{x_1},{x_2},{x_3}} \right) = \left( {{x_3} - {x_2}} \right){{\bf{b}}_1} - \left( {{x_1} - {x_3}} \right){{\bf{b}}_2} + \left( {{x_1} - {x_2}} \right){{\bf{b}}_3}\)

  1. Compute\(T\left( {{{\bf{e}}_1}} \right)\), \(T\left( {{{\bf{e}}_2}} \right)\) and \(T\left( {{{\bf{e}}_3}} \right)\).
  2. Compute \({\left( {T\left( {{{\bf{e}}_1}} \right)} \right)_B}\), \({\left( {T\left( {{{\bf{e}}_2}} \right)} \right)_B}\) and \({\left( {T\left( {{{\bf{e}}_3}} \right)} \right)_B}\).
  3. Find the matrix for \(T\) relative to \(\varepsilon \), and\(B\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{.4}&{ - .3}\\{.4}&{1.2}\end{aligned}} \right)\). Explain why \({A^k}\) approaches \(\left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\) as \(k \to \infty \).

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.
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