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Question 20: Use a property of determinants to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

Short Answer

Expert verified

It is proved that \(A\) and \({A^T}\) have the same characteristic polynomial.

Step by step solution

01

Properties of determinants

Consider \(A\) and \(B\) as \(n \times n\) matrices.

  1. \(A\)isinvertiblesuch that \(\det A \ne 0\).
  2. \(\det AB = \left( {\det A} \right)\left( {\det B} \right)\)
  3. \(\det {A^T} = \det A\)
  4. When \(A\) is triangular, the product of the entries on the main diagonal of \(A\) is \(\det A\).
02

Show that \(A\) and \({A^T}\) have the same characteristic polynomial

Use the property of determinant to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

\[\begin{array}\det \left( {{A^T} - \lambda I} \right) = \det \left( {{A^T} - \lambda {I^T}} \right)\\ = \det {\left( {A - \lambda I} \right)^T}{\rm{ }}\left( {{\mathop{\rm by}\nolimits} \,\,{\mathop{\rm transpose}\nolimits} \,\,{\mathop{\rm property}\nolimits} } \right)\\ = \det \left( {A - \lambda I} \right)\end{array}\]

Thus, it is proved that \(A\) and \({A^T}\) have the same characteristic polynomial.

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

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

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

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.

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