Linear Algebra

Section II. Linear Geometry 41Proof (We’ll use some algebraic properties of dot product that we have not yetchecked, for instance that ⃗u • (⃗a + ⃗b) = ⃗u • ⃗a + ⃗u • ⃗b and that ⃗u • ⃗v = ⃗v • ⃗u. SeeExercise 18.) Since all the numbers are positive, the inequality holds if and onlyif its square holds.‖⃗u + ⃗v ‖ 2 ( ‖⃗u ‖ + ‖⃗v ‖ ) 2( ⃗u + ⃗v ) • ( ⃗u + ⃗v ) ‖⃗u ‖ 2 + 2 ‖⃗u ‖ ‖⃗v ‖ + ‖⃗v ‖ 2⃗u • ⃗u + ⃗u • ⃗v + ⃗v • ⃗u + ⃗v • ⃗v ⃗u • ⃗u + 2 ‖⃗u ‖ ‖⃗v ‖ + ⃗v • ⃗v2 ⃗u • ⃗v 2 ‖⃗u ‖ ‖⃗v ‖That, in turn, holds if and only if the relationship obtained by multiplying bothsides by the nonnegative numbers ‖⃗u ‖ and ‖⃗v ‖and rewriting2 ( ‖⃗v ‖ ⃗u ) • ( ‖⃗u ‖⃗v ) 2 ‖⃗u ‖ 2 ‖⃗v ‖ 20 ‖⃗u ‖ 2 ‖⃗v ‖ 2 − 2 ( ‖⃗v ‖ ⃗u ) • ( ‖⃗u ‖⃗v ) + ‖⃗u ‖ 2 ‖⃗v ‖ 2is true. But factoring shows that it is true0 ( ‖⃗u ‖⃗v − ‖⃗v ‖ ⃗u ) • ( ‖⃗u ‖⃗v − ‖⃗v ‖ ⃗u )since it only says that the square of the length of the vector ‖⃗u ‖⃗v − ‖⃗v ‖ ⃗u isnot negative. As for equality, it holds when, and only when, ‖⃗u ‖⃗v − ‖⃗v ‖ ⃗u is⃗0. The check that ‖⃗u ‖⃗v = ‖⃗v ‖ ⃗u if and only if one vector is a nonnegative realscalar multiple of the other is easy.QEDThis result supports the intuition that even in higher-dimensional spaces,lines are straight and planes are flat. We can easily check from the definitionthat linear surfaces have the property that for any two points in that surface,the line segment between them is contained in that surface. But if the linearsurface were not flat then that would allow for a shortcut.PQBecause the Triangle Inequality says that in any R n the shortest cut betweentwo endpoints is simply the line segment connecting them, linear surfaces haveno bends.Back to the definition of angle measure. The heart of the Triangle Inequality’sproof is the ⃗u • ⃗v ‖⃗u ‖ ‖⃗v ‖ line. We might wonder if some pairs of vectorssatisfy the inequality in this way: while ⃗u • ⃗v is a large number, with absolutevalue bigger than the right-hand side, it is a negative large number. The nextresult says that does not happen.