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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

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 \(p\left( \lambda \right)\).

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

See the step by step solution

Step by Step Solution

Step 1: 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.

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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