Advanced book on Mathematics Olympiad

Algebra 373

∥

∥∥

∥∥

(m

∏

i= 1

Vi

)

x−

(m

∏

i= 1

Wi

)

x

∥

∥∥

∥∥=‖VVmx−WWmx‖

=‖V(Vm−Wm)x+(V−W)Wmx‖.

Now we use the triangle inequality to increase the value of this expression to

‖V(Vm−Wm)x‖+‖(V−W)Wmx‖.

From the fact thatVis an isometry it follows that

‖V(Vm−Wm)x‖=‖(Vm−Wm)x‖≤ 1.

From the fact thatWmis an isometry, it follows that‖Wmx‖≤1, and so‖(V−W)Wmx‖≤
m−1 by the induction hypothesis. Putting together the two inequalities completes the
induction, and the inequality is proved.

Remark.In quantum mechanics the vector spaces are complex (not real) and the word
isometryis replaced byunitary. Unitary linear transformations model evolution, and the
above property shows that (measurement) errors accumulate linearly.

119.Place triangleABCin the complex plane such that the coordinates of the verticesA,
B, andCare, respectively, the third roots of unity 1,,^2. Callzthe complex coordinate
ofP. Start with the obvious identity

(z− 1 )+(z−)+^2 (z−^2 )= 0.

Move one term to the other side:

−^2 (z−^2 )=(z− 1 )+(z−).

Now take the absolute value and use the triangle inequality:

|z−^2 |=|(z− 1 )+(z−)|≤|z− 1 |+|((z−)|=|z− 1 |+|z−^2 |.

Geometrically, this isPC≤PA+PB.
Equality corresponds to the equality case in the triangle inequality for complex num-
bers, which holds if the complex numbers have positive ratio. Specifically,(z− 1 )=
a(z−)for some positive real numbera, which is equivalent to

z− 1

z−

=a.

In geometric terms this means thatPAandPBform an angle of 120◦, so thatPis on

the arc



AB. The other two inequalities are obtained by permuting the letters.

(D. Pompeiu)