By Gerard G. Emch

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**Example text**

17) for F or for the matrix of F. Hint. For 2 example, the matrix of R will equal (4 + K2) -1/2 ( -K K 2 0 ) 0 , o 0 (4+K2)1/2 and so forth, but you should attempt to compute such results directly from F. 8. Show that QAQT E Psym for every A E Psym and every Q E Orth. 21) where the square root here is the positive root. Hint. 2). Alternatively, an indirect proof can be based on the uniqueness of the positive square root. 9. 22) and for all vectors w, u, v E V in the case dim V = 3. Hint. Let {el, e2, e3} be a fixed ON right-handed basis with (e 1,e 2 x e 3) = +1, set ri := Rei for i = 1,2,3, and show that {r},r2, r3} is also an ON basis and R = (ri®ri)R = r i ® (RT ri) = r i ® e i .

6) so the system can be solved to give 1) I) aa a~ -1 ( A. 21). 21) in the case n = 3. An analogous proof is given in TRUESDELL & NOLL (1965) in the general case of an arbitrary positive integer n. 1. 4) for the principal invariants in the case n = 3. 2. 17). Hint. Calculate RART = (2f®f - I)A(2f®f - I) where f is an eigenvector of A (Af = af for a corresponding eigenvalue a). You should find RART = A from which the desired result will follow. S. Let v be a fixed point (Rv = v) of the rotation R = 2f0f - I, where f is a given fixed unit vector.

In(A) are called the principal invariants of A. 5) For any Q E Orth, one finds directly the results det (AI + QAQT) = det (Q(AI + A)QT) = det(AI + A), where the last equality follows from the product rule for determinants along with the result 1 = det I = det QQT = (det Q) (det QT) for every orthogonal Q. 3) for A is the same as that for QAQT, from which one has the result Ii (QAQT) = Ii(A) for j = 1,2, ... 6) for every Q E Orth and for every A E Lin. Hence the principal invariants Ii = Ii(A) are isotropic scalar-valued functions.