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Chapter 5: Q17SE (page 267)

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

Short Answer

Expert verified

It is proved that \(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). It is also proved that the eigenvalues of a \(2 \times 2\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).

Step by step solution

01

Step 1: Find the characteristic polynomial.

Determine\(\det \left( {A - \lambda I} \right)\).

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) &= \det \left( {\begin{aligned}{*{20}{c}}{{a_{11}} - \lambda }&{{a_{12}}}\\{{a_{21}}}&{{a_{22}} - \lambda }\end{aligned}} \right)\\ &= \left( {{a_{11}} - \lambda } \right)\left( {{a_{22}} - \lambda } \right) - {a_{12}}{a_{21}}\\ &= {\lambda ^2} - {a_{11}}\lambda - {a_{22}}\lambda + {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\\ &= {\lambda ^2} - \left( {{a_{11}} + {a_{22}}} \right)\lambda + \left( {{a_{11}}{a_{22}} - {a_{12}}{a_{21}}} \right)\end{aligned}\)

Now use a trace of \(A\), that is, \({\rm{tr}}\;A = {a_{11}} + {a_{22}}\) then we have,

\(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {trA} \right)\lambda + \det A\)

Hence it is proved that\(\det \left( {A - \lambda I} \right) = {\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\).

02

Step 2: Show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\)

Determine the eigenvalues of \(A\).

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) &= 0\\{\lambda ^2} - \left( {{\rm{tr}}\;A} \right)\lambda + \det A &= 0\\\lambda &= \frac{{tr\;A \pm \sqrt {{{\left( {tr\;A} \right)}^2} - 4\det A} }}{2}\end{aligned}\)

As eigenvalues are positive if the discriminant is positive.

\(\begin{aligned}{c}{\left( {{\rm{tr}}A} \right)^2} - 4\det A \ge 0\\{\left( {{\rm{tr}}A} \right)^2} \ge 4\det A\\\frac{{{{\left( {{\rm{tr}}A} \right)}^2}}}{4} \ge \det A\\{\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2} \ge \det A\end{aligned}\)

Thus, \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).

Hence it is proved that the eigenvalues of a \(2 \times 2\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{trA}}{2}} \right)^2}\).

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Most popular questions from this chapter

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.

Exercises 19鈥�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 \(p\).

a. Let \(A\) be a diagonalizable \(n \times n\) matrix. Show that if the multiplicity of an eigenvalue \(\lambda \) is \(n\), then \(A = \lambda I\).

b. Use part (a) to show that the matrix \(A =\left({\begin{aligned}{*{20}{l}}3&1\\0&3\end{aligned}}\right)\) is not diagonalizable.

Let\(T:{{\rm P}_2} \to {{\rm P}_3}\) be a linear transformation that maps a polynomial \({\bf{p}}\left( t \right)\) into the polynomial \(\left( {t + 5} \right){\bf{p}}\left( t \right)\).

  1. Find the image of\({\bf{p}}\left( t \right) = 2 - t + {t^2}\).
  2. Show that \(T\) is a linear transformation.
  3. Find the matrix for \(T\) relative to the bases \(\left\{ {1,t,{t^2}} \right\}\) and \(\left\{ {1,t,{t^2},{t^3}} \right\}\).

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.
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