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Q5SE

Expert-verifiedFound in: Page 267

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

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

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.

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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