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Chapter 5: Q5.4-31E (page 267)

Find the \(B\) matrix for the transformation\({\rm{x}} \mapsto A{\rm{x}}\), when \(B = \left\{ {{b_1},{b_2},{b_3}} \right\}\).

\(A = \left( {\begin{aligned}{}{ - 7}&{ - 48}&{ - 16}\\1&{14}&6\\{ - 3}&{ - 45}&{ - 19}\end{aligned}} \right)\),

\({b_1} = \left( {\begin{aligned}{}{ - 3}\\1\\{ - 3}\end{aligned}} \right)\), \({b_2} = \left( {\begin{aligned}{}{ - 2}\\1\\{ - 3}\end{aligned}} \right)\), \({b_3} = \left( {\begin{aligned}{}{\,\,\,3}\\{ - 1}\\{\,\,\,0}\end{aligned}} \right)\)

Short Answer

Expert verified

The B matrix of the transformation \({\rm{x}} \mapsto A{\rm{x}}\)is \(D = {P^{ - 1}}AP\), which is \(\left( {\begin{aligned}{}{ - 7}&{}&{ - 2}&{}&{ - 6}\\0&{}&{ - 4}&{}&{ - 6}\\0&{}&0&{}&{ - 1}\end{aligned}} \right)\)

Step by step solution

01

Find P matrix

Let \(P\) is an invertible matrix and \(D\) is the diagonal matrix, such that the matrix A can be written as follows:

\(A = PD{P^{ - 1}}\)

Where the columns of the \(P\) matrix are the same as that of basis \(B\). So, the matrix \(P\) can be written as:

\(\begin{aligned}{c}P &= \left( {{{\rm{b}}_1}\,\,{{\rm{b}}_2}\,\,{{\rm{b}}_3}} \right)\\ &= \left( {\begin{aligned}{}{ - 3}&{}&{ - 2}&{}&3\\1&{}&1&{}&{ - 1}\\{ - 3}&{}&{ - 3}&{}&0\end{aligned}} \right)\end{aligned}\)

02

Find D matrix

As \(A = PD{P^{ - 1}}\) then the diagonal matrix \(D\) can be obtained as \(D = {P^{ - 1}}AP\). To find \(D = {P^{ - 1}}AP\), first, find \({P^{ - 1}}\)using the cofactors and determinant method.

\(\begin{aligned}{c}P &= \left( {\begin{aligned}{}{ - 3}&{}&{ - 2}&{}&3\\1&{}&1&{}&{ - 1}\\{ - 3}&{}&{ - 3}&{}&0\end{aligned}} \right)\\{P^{ - 1}} &= \left( {\begin{aligned}{}{ - 1}&{}&{ - 3}&{}&{ - 1/3}\\{\,\,1}&{}&{\,\,3}&{}&{\,0}\\{\,\,0}&{}&{ - 1}&{}&{ - 1/3}\end{aligned}} \right)\end{aligned}\)

Now find the matrix \(D\) as follows:

\(\begin{aligned}{c}D &= {P^{ - 1}}AP\\ &= \left( {\begin{aligned}{}{ - 1}&{}&{ - 3}&{}&{ - 1/3}\\{\,\,1}&{}&{\,\,3}&{}&{\,0}\\{\,\,0}&{}&{ - 1}&{}&{ - 1/3}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 7}&{}&{ - 48}&{}&{ - 16}\\1&{}&{14}&{}&6\\{ - 3}&{}&{ - 45}&{}&{ - 19}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 3}&{}&{ - 2}&{}&3\\1&{}&1&{}&{ - 1}\\{ - 3}&{}&{ - 3}&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}{ - 7}&{}&{ - 2}&{}&{ - 6}\\0&{}&{ - 4}&{}&{ - 6}\\0&{}&0&{}&{ - 1}\end{aligned}} \right)\end{aligned}\)

The B matrix of the transformation \({\rm{x}} \mapsto A{\rm{x}}\) is \(D = {P^{ - 1}}AP\), which is \(\left( {\begin{aligned}{}{ - 7}&{}&{ - 2}&{}&{ - 6}\\0&{}&{ - 4}&{}&{ - 6}\\0&{}&0&{}&{ - 1}\end{aligned}} \right)\).

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Most popular questions from this chapter

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 \(B = \left\{ {1,t,{t^2}} \right\}\).

[M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

20. \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{2}}&{{\bf{12}}}&{{\bf{13}}}&{{\bf{11}}}\\{{\bf{ - 2}}}&{\bf{3}}&{\bf{0}}&{\bf{2}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}&{\bf{2}}\end{array}} \right]\]

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

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

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)
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