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Chapter 5: Q5SE (page 267)

If \(p\left( t \right) = {c_0} + {c_1}t + {c_2}{t^2} + ...... + {c_n}{t^n}\), define \(p\left( A \right)\) to be the matrix formed by replacing each power of \(t\) in \(p\left( t \right)\)by the corresponding power of \(A\) (with \({A^0} = I\) ). That is,

\(p\left( t \right) = {c_0} + {c_1}I + {c_2}{I^2} + ...... + {c_n}{I^n}\)

Show that if \(\lambda \) is an eigenvalue of A, then one eigenvalue of \(p\left( A \right)\) is\(p\left( \lambda \right)\).

Short Answer

Expert verified

It is proved that if \(\lambda \) is an eigenvalue of \(\lambda \), then one eigenvalue of \(p\left( A \right)\) is \(p\left( \lambda \right)\).

Step by step solution

01

Definition of eigenvalue

Eigenvalues are a special set of scalars associated with a linear system of equations that are sometimes also known as characteristic roots, characteristic values, proper values, or latent roots.

02

Step 2: Showing if is an eigenvalue of \(A\), then one eigenvalue of \(p\left( A \right)\) is \(p\left( \lambda  \right)\)

Suppose that\(\lambda \)is an eigenvalue of\(A\)with the corresponding eigenvector\({\rm{x}}\), that is, suppose that\(A{\rm{x}} = \lambda {\rm{x}},{\rm{x}} \ne 0\).

Then,

\(\begin{array}{r}p\left( A \right){\rm{x}} = \left( {{c_0}I + {c_1}A + \ldots + {c_n}{A^n}} \right){\rm{x}}\\ = {c_0}{\rm{x}} + {c_1}A{\rm{x}} + \ldots + {c_n}{A^n}{\rm{x}}\end{array}\)

From exercise 4 we know that\({A^n}{\rm{x}} = {\lambda ^n}{\rm{x}}\), so:

\(\begin{array}{c}p\left( A \right){\rm{x}} = {c_0}{\rm{x}} + {c_1}\lambda {\rm{x}} + \ldots + {c_n}{\lambda ^n}{\rm{x}}\\ = \left( {{c_0} + {c_1}\lambda + \ldots + {c_n}{\lambda ^n}} \right){\rm{x}}\\ = p(\lambda ){\rm{x}}\end{array}\).

Hence it is proved that \(\lambda \) is an eigenvalue of \(\lambda \), then one eigenvalue of \(p\left( A \right)\) is \(p\left( \lambda \right)\).

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Most popular questions from this chapter

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

9. \(\left( {\begin{array}{*{20}{c}}3&{ - 1}\\1&5\end{array}} \right)\)

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

Let\(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\) and \(D = \left\{ {{{\bf{d}}_{\bf{1}}},{{\bf{d}}_{\bf{2}}}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\)

Find the matrix for \(T\) relative to \(B\), and\(D\).

For the Matrices A find real closed formulas for the trajectory x鈫�(t+1)=Ax鈫�(t)where x(0)鈫�=[01]

A=[43-34]

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

16. \(\left( {\begin{array}{*{20}{c}}{\bf{0}}&{{\bf{ - 4}}}&{{\bf{ - 6}}}\\{{\bf{ - 1}}}&{\bf{0}}&{{\bf{ - 3}}}\\{\bf{1}}&{\bf{2}}&{\bf{5}}\end{array}} \right)\)
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