1916-17.] The Square Roots of a Linear Vector Function. 353 
The reasoning fails and the solution is impossible if g = 0. As an 
example, we cannot have cp such that cp 2 =jSip + kSj p. 
When m is of Class IV, i.e. reducible, we may write 
wp = 9P + yS Xp ...... (4) 
where the axes of co are y and any vector perpendicular to X. Let a be 
now selected any vector perpendicular to X. Let us assign cpa = /3, where 
/3 is a different vector perpendicular to X. We now have 
<p/3 = <p 2 a = too. = ga , 
that is, cp is involutory as regards the directions of a and /3. If y is not 
perpendicular to X it is an axis of cp (assuming that y and X are actual 
vectors, i.e. that co is not a mere dilation). For if y is not an axis of cp , 
both y and cpy are axes of co having the same root, and co has two different 
planes of axes, which is impossible. Hence <py is determined except for 
sign. Since <pa is arbitrary within the plane SX/o = 0, we may construct 
an infinite number of different functions cp satisfying (1). When so con- 
structed the latent roots of (p are, besides that corresponding to y, the 
two square roots of g, opposite in sign. Or we may begin by choosing 
any two vectors perpendicular to X to be axes of <p , with roots ± Jg, 
respectively. On the other hand, if we assign the same value of Jg to 
two axes of <p , cp becomes reducible and has only two forms, corresponding 
to choice of signs for cpy. 
It remains to consider the case where SyX = 0, which includes the 
problem of finding the square root of a shear. In the plane of axes all 
is as before, cp is then fully determined when we know cp\. Assume r 
selected as + Jg with p and v axes of cp in the plane SXp = 0, so that 
cpp = rp and cpv= — rv. Also assume (p\ = x\ + yp +zv. Then 
<p 2 \ — x<p\ + ypp + zpv 
= + xyp + xzv 
+ ryp — rzv 
— oj/V 
-=g\ +yX 2 . 
Expanding yX 2 in terms of p and v, we may write yX 2 — ap + bv where 
a and b are known. Equating coefficients of X, p, v, we have 
x 2, = g , y(x + r) — a , z(x - r) = b. 
Since x= ±r, it follows that a or b must be zero. Hence if X is real, p or v 
must coincide in direction with y, the double axis of w. We then have 
either y or z arbitrary. The solution again fails when g = 0 (unless X 2 = 0) 
YOL. XXXVII. 23 
