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Linear Algebra and its Applications
Found in: Page 267
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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.

It is proved that \(p\left( A \right)\) is the zero matrix.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition of Caylee Hamilton theorem

This theorem states that every square matrix satisfies the characteristic equation of its own.

If there is a square matrix \(A\) we can say that the matrix satisfies the following equation \(p\left( \lambda \right) = \det \left( {A - \lambda I} \right)\). Then we have \(p\left( A \right) = 0\).

Step 2: Defining \(p\left( A \right)\) 

First of all, note that if \(A = PD{P^{ - 1}}\), then for any natural \(k\).

\(\begin{aligned}{c}{A^k} &= A \cdot A \cdot A \cdot \ldots \cdot A\\ &= \left( {PD{P^{ - 1}}} \right)\left( {PD{P^{ - 1}}} \right)\left( {PD{P^{ - 1}}} \right) \ldots \left( {PD{P^{ - 1}}} \right)\\ &= PD\left( {{P^{ - 1}}P} \right)D\left( {{P^{ - 1}}P} \right)D\left( {{P^{ - 1}} \cdot \ldots \cdot P} \right)D{P^{ - 1}}\\ &= P{D^k}{P^{ - 1}}\end{aligned}\)\(\)

Step 3: Showing that \(p\left( A \right)\) is the zero matrix

Write \(p\left( A \right)\).

\(\begin{aligned}{c}p\left( A \right) &= {c_0}I + {c_1}A + {c_2}{A^2} + \ldots + {c_n}{A^n}\\ &= {c_0}PI{P^{ - 1}} + {c_1}PD{P^{ - 1}} + {c_2}P{D^2}{P^{ - 1}} + \ldots + {c_n}P{D^n}{P^{ - 1}}\\ &= P\left( {{c_0}I + {c_1}D + {c_2}{D^2} + \ldots + {c_n}{D^n}} \right){P^{ - 1}}\end{aligned}\)

Since \(I,D,{D^2}, \ldots ,{D^n}\) are diagonal matrices, the matrix in the middle is also diagonal as a linear combination of diagonal matrices.

The \(k\)-th entry of this matrix is \({c_0} + {c_1}{\lambda _k} + {c_2}\lambda _k^2 + \ldots + {c_n}\lambda _k^n = p\left( {{\lambda _k}} \right)\), where \({\lambda _k}\) is a \(k\)-th eigenvalue of \(A\).

But \(p\left( t \right)\) is the characteristic polynomial of \(A\), so for any eigenvalue \({\lambda _k}\) of \(A\) we have \(p\left( {{\lambda _k}} \right) = 0\).

It follows that all the entries of the matrix \({c_0}I + {c_1}D + {c_2}{D^2} + \ldots + {c_n}{D^n}\) are zeros, and then

\(\begin{aligned}{c}p\left( A \right) &= P \cdot 0 \cdot {P^{ - 1}}\\ &= 0\end{aligned}\)

It is proved that \(p\left( A \right)\) is the zero matrix.

Most popular questions for Math Textbooks

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

For the matrices A find real closed formulas for the trajectory x→(t+1)=Ax→(t) where x→(0)=[01]. Draw a rough sketch A=[1-31.2-2.6]

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

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

Question: Is \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\)? If so, find the eigenvalue.

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\(B\).
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