Utilizing Cayley-Hamilton's Theorem to Solve N x N Determinant Problem

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The discussion focuses on utilizing Cayley-Hamilton's Theorem to demonstrate the non-negativity of specific determinants involving matrices A and B. Participants explore the expressions det(A^2 + A + E) and det(E + A + B + A^2 + B^2), with suggestions to manipulate the terms for clearer results. There is a consensus that separating cases based on whether det(A) equals zero or not allows for more flexible handling of the matrix A. One participant identifies a potential solution involving the Cayley-Hamilton context, indicating progress in the problem-solving process. The conversation emphasizes the importance of selecting appropriate terms to simplify the determinant calculations.
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1. Given A,B\in Mat _n(\mathbb{R})

2. Show that:
a) \det (A^2 + A + E)\geq 0
b) \det (E+A+B+A^2+B^2)\geq 0 ,
where E is the unit matrix.
3. My attempt at a solution
A^2 + A + E=(A + E)^2 -2A


https://drive.google.com/file/d/0B8zKPTh1siSsOHNWQnBfaXR3QXM/view?usp=sharing
Snapshot.jpg

pleas give me tips to solve
 
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kockabogyo said:
A^2 + A + E=(A + E)^2 -2A
Something went wrong with the linear term, and I would choose a different term to square.

You can consider the cases ##det(A)=0## and ##det(A) \neq 0## separately, that gives more freedom to manipulate A in one case.
 
mfb said:
Something went wrong with the linear term, and I would choose a different term to square.

You can consider the cases ##det(A)=0## and ##det(A) \neq 0## separately, that gives more freedom to manipulate A in one case.

Thanks, yes, sorry not - 2A only -A , but than?
 
I would choose a different term to square. A term that doesn't leave an A outside.
 
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mfb said:
I would choose a different term to square. A term that doesn't leave an A outside.

O Yeah!.. I think I found it.. Cayley Hamilton's context A2 - Tr(A)*A+det(A)*E = O
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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