StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q7SE

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

**Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.**

** **

It is proved that \(p\left( A \right)\) is the zero matrix.

This theorem states that every square matrix satisfies the characteristic equation of 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.

94% of StudySmarter users get better grades.

Sign up for free