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Found in: Page 267

### 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

# 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.

See the step by step solution

## Step 1: Definition of Caylee Hamilton theorem

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 $$p\left( A \right) = 0$$.

## Step 2: Defining $$p\left( A \right)$$

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}

## Step 3: Showing that $$p\left( A \right)$$ is the zero matrix

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.