Can every symmetric matrix be a matrix of inertia?

AI Thread Summary
Any real 3x3 symmetric matrix can be diagonalized using orthogonal matrices, regardless of the signs of its eigenvalues. However, for a matrix to represent a realistic matrix of inertia, it must have positive eigenvalues, which implies that the diagonal elements should also be positive. Off-diagonal elements can influence the matrix's properties, but they do not necessarily have to be positive. Negative entries are generally not suitable for a matrix of inertia, particularly on the diagonal. Therefore, while any symmetric matrix can be diagonalized, not all can serve as a valid matrix of inertia.
sfn17
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Hello,

I am often designing math exams for students of engineering.

What I ask is the following:

Can I choose any real 3x3 symmetric matrix with positive eigenvalues as a realistic matrix of inertia?

Possibly, there are secret connections between the off-diagonal elements (if not zero) which I should have in mind...

I am not quite sure that the off-diagonal elements can be negative. Is it possible to have negative entries even on the diagonal line?

sfn17
 
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We have mathematical theorem that any real number symmetric matrix, with no condition on signs of its eigenvalues, can be diagonarized by coordinate transformation via orthogonal matrices.
 
I have Mass A being pulled vertically. I have Mass B on an incline that is pulling Mass A. There is a 2:1 pulley between them. The math I'm using is: FA = MA / 2 = ? t-force MB * SIN(of the incline degree) = ? If MB is greater then FA, it pulls FA up as MB moves down the incline. BUT... If I reverse the 2:1 pulley. Then the math changes to... FA = MA * 2 = ? t-force MB * SIN(of the incline degree) = ? If FA is greater then MB, it pulls MB up the incline as FA moves down. It's confusing...

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