Solutions To (Relatively) Simpler Tensor Algebra Problems

 Solutions:

a) We have:  I = K x 

(1….0…..0)

(0….1…..1)

(0….1… ..1)

 We write out the determinant with eigenvalue  l:

(1 - l….0…..0)

(0….1 - l   ..1)

(0….1… ..1 - l)

Leading to the characteristic equation:

(1 - l)³ – (1 - l) = 0

Factoring:

(1 - l) [ ((1 - l)² – 1] = 0 

Or:

(1 - l) (l² –  2l) = 0

Yielding eigenvalues: l= 0, l = 2

Then:

T = Kl,  so:  T1 = 0, T2 = K, and T3 =2K

Or: (0, K, 2K)  for the principal moments of inertia.

b) We have to take:  (I – T1)C

So that we obtain:

=

 (0….0…..0) (x)

  (0….0…..1) (y)

  (0….1… ..0) (z)   =   0

 Finally:  0  =

(0)

(z)

(y)  

 Where ‘0’ for the x  element in column matrix implies the direction is i.

c) By the analog of the parallel axis theorem:

I _o = I_G -   M(R² I – RR) 

D I =  I _o -  I_G   =    M(R² I – RR)

RR =   

And:

D I =   

Finally:  D I =