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

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