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

kockabogyo · Nov 3, 2015

1. Given [itex]A,B\in Mat _n(\mathbb{R})[/itex]

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

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

Handwritten-style typeset computation: A is the 2 by 2 matrix [a b; c d], A squared equals [a^2+b*c, a*b+b*d; c*a+d*c, c*b+d^2], E is the identity matrix [1 0; 0 1], and A^2+A+E is written out entrywise, then regrouped as [a^2+b*c-a+1, b*(a-d+1); c*(a+d+1), c*b+d^2+d+1].

pleas give me tips to solve

mfb · Nov 3, 2015

kockabogyo said:

[itex]A^2 + A + E[/itex]=[itex](A + E)^2 -2A[/itex]

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.

kockabogyo · Nov 3, 2015

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?

mfb · Nov 3, 2015

I would choose a different term to square. A term that doesn't leave an A outside.

kockabogyo · Nov 3, 2015

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

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

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