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Chapter 5: Q5.523E (page 267)

Chapter 7 will focus on matricesAwith the property that\({A^T} = A\).
Exercises 23 and 24 show that every eigenvalue of such a matrix
is necessarily real.

Let \(A\) be an \(n \times n\) real matrix with the property that \({A^T} = A\), let \({\bf{x}}\) be any vector in \({\mathbb{C}^n}\), and let \(q = {\overline {\bf{x}} ^{\bar T}}A{\bf{x}}\). The equalities below show that \(q\) is a real number by verifying that \(\bar q = q\). Give a reason for each step.

\(\begin{aligned}{}\bar q = \overline {{{\bar x}^T}Ax} = {x^T}\overline {Ax} = {x^T}A\bar x = {\left( {{x^T}A\bar x} \right)^T} = {{\bar x}^T}{A^T}x = q\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( a \right)\,\,\,\,\,\,\,\,\,\,\,\left( b \right)\,\,\,\,\,\,\,\,\,\,\,\left( c \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( d \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( e \right)\end{aligned}\)

Short Answer

Expert verified

All steps have been explained in proving \(\bar q = q\).

Step by step solution

01

Definition of the matrix.

A rectangular aligned of numbers, symbols, or expressions, arranged in rows and columns.

02

Verifying \(\bar q = q\) .

  1. It is known from the properties of conjugates and transposes that \({\overline {\bar x} ^T} = {x^T}\)and \(\overline {AB} = \bar A\bar B\). Therefore, \(\overline {{{\bar x}^T}Ax} = {x^T}\overline {Ax} \).
  1. \(\bar A = A\), because \(A\) is real. Therefore, \(\bar A = A\). So, \({x^T}\overline {Ax} = {x^T}A\bar x\).
  1. \({x^T}A\bar x\)is a scalar. So, the transpose does not change it.
  1. \({(AB)^T} = {B^T}{A^T}\)- property of transpose operator.
  1. \({A^T} = A\) and definition of \(q\).

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Most popular questions from this chapter

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

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

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

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)
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