Let A be an n xn matrix and f(r) = Cr* + ... + cr + co, a polynomial. By f(A) we mean the n x n matrix Ck A* + ck-1Ak-1+….+cqA+ coIn•. Prove if A is an nxn matrix then there exists a polynomial f of degree at most n2 such that f(A) = 0,xn- Note you cannot simply quote a theorem like Cayley-Hamilton, you have to prove it yourself.
Let A be an n xn matrix and f(r) = Cr* + ... + cr + co, a polynomial. By f(A) we mean the n x n matrix Ck A* + ck-1Ak-1+….+cqA+ coIn•. Prove if A is an nxn matrix then there exists a polynomial f of degree at most n2 such that f(A) = 0,xn- Note you cannot simply quote a theorem like Cayley-Hamilton, you have to prove it yourself.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
Let \( A \) be an \( n \times n \) matrix and \( f(x) = c_k x^k + \dots + c_1 x + c_0 \), a polynomial. By \( f(A) \) we mean the \( n \times n \) matrix:
\[ f(A) = c_k A^k + c_{k-1} A^{k-1} + \cdots + c_1 A + c_0 I_n. \]
Prove that if \( A \) is an \( n \times n \) matrix then there exists a polynomial \( f \) of degree at most \( n^2 \) such that \( f(A) = 0_{n \times n} \).
Note: you cannot simply quote a theorem like Cayley-Hamilton, you have to prove it yourself.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc854bee8-14af-49f6-8ae2-48b2949ea5fc%2Ff4b517dc-f752-44ee-9d99-4eceefcd536d%2Fz3arvbl_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Let \( A \) be an \( n \times n \) matrix and \( f(x) = c_k x^k + \dots + c_1 x + c_0 \), a polynomial. By \( f(A) \) we mean the \( n \times n \) matrix:
\[ f(A) = c_k A^k + c_{k-1} A^{k-1} + \cdots + c_1 A + c_0 I_n. \]
Prove that if \( A \) is an \( n \times n \) matrix then there exists a polynomial \( f \) of degree at most \( n^2 \) such that \( f(A) = 0_{n \times n} \).
Note: you cannot simply quote a theorem like Cayley-Hamilton, you have to prove it yourself.
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