91Ó°ÊÓ

This theorem states that every square matrix satisfies the characteristicequationof 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 \(p\left( A \right) = 0\).

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

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

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.