MATH 304Linear AlgebraLecture 16b:Euclidean structure in Rn.Vectors: geometric approachABA′B′• A vector is represented by a directed segment.• Directed segment is drawn as an arrow.• Different arrows represent the same vector ifthey are of the same length and direction.Vectors: geometric approachABA′B′vv−→AB denotes the vector represented by the arrowwith tip at B and tail at A.−→AA is called the zero vector and denoted 0.Vectors: geometric approachABA′B′−vvIf v =−→AB then−→BA is called the negative vector ofv and denoted −v.Vector additionGiven vectors a and b, their sum a + b is defined bythe rule−→AB +−→BC =−→AC .That is, choose points A, B, C so that−→AB = a and−→BC = b. Then a + b =−→AC .ABCA′B′C′aba + baba + bThe difference of the two vectors is defined asa − b = a + (−b).a − bbaScalar multiplicationLet v be a vector and r ∈ R. By definition, rv is avector whose magnitude is |r| times the magnitudeof v. The direction of r v coincides with that of v ifr > 0. If r < 0 then the directions of rv and v areopposite.v3v−2vBeyond linearity: length of a vectorThe length (or the magnitude) of a vector−→AB isthe length of the representing segment AB. Thelength of a vector v is denoted |v| or kvk.Properties of vector length:|x| ≥ 0, |x| = 0 only if x = 0 (positivity)|rx| = |r ||x| (homogeneity)|x + y| ≤ |x| + |y| (triangle inequality)xyx + yBeyond linearity: angle between vectorsGiven nonzero vectors x and y, let A, B, and C bepoints such that−→AB = x and−→AC = y. Then ∠BACis called the angle between x and y.The vectors x and y are called orthogonal (denotedx ⊥ y) if the angle between them equals 90o.A BCθyxxx + yyPythagorean Theorem:x ⊥ y =⇒ |x + y|2= |x|2+ |y|23-dimensional Pythagorean Theorem:If vectors x, y, z are pairwise orthogonal then|x + y + z|2= |x|2+ |y|2+ |z|2A BCθyxx − yLaw of cosines:|x − y|2= |x|2+ |y|2− 2|x||y| cos θBeyond linearity: dot productThe dot product of vectors x and y isx · y = |x||y| cos θ,where θ is the angle between x and y.The dot product is also called the scalar product.Alternative notation: (x, y) or hx, yi.The vectors x and y are orthogonal if and only ifx · y = 0.Relations between lengths and dot products:• |x| =√x · x• |x · y| ≤ |x||y|• |x − y|2= |x|2+ |y|2− 2 x·yVectors: algebraic approachAn n-dimensional co ordinate vector is an element ofRn, i.e., an ordered n-tuple (x1, x2, . . . , xn) of realnumbers.Let a = (a1, a2, . . . , an) and b = (b1, b2, . . . , bn) bevectors, and r ∈ R be a scalar. Then, by definition,a + b = (a1+ b1, a2+ b2, . . . , an+ bn),ra = (ra1, ra2, . . . , ran),0 = (0, 0, . . . , 0),−b = (−b1, −b2, . . . , −bn),a − b = a + (−b) = (a1− b1, a2− b2, . . . , an− bn).Cartesian coordinates: geometric meets algebraic(−3, 2)(2, 1)(−3, 2)(2, 1)Once we specify an origin O, each point A isassociated a position vector−→OA. Conversely, everyvector has a unique representative with tail at O.Cartesian coordinates allow us to identify a line, aplane, and space with R, R2, and R3, respectively.Length and distanceDefinition. The length of a vectorv = (v1, v2, . . . , vn) ∈ Rniskvk =pv21+ v22+ ··· + v2n.The distance between vectors/points x and y isky − xk.Properties of length:kxk ≥ 0, kxk = 0 only if x = 0 (positivity)krxk = |r |kxk (homogeneity)kx + yk ≤ kxk + kyk (triangle inequality)Scalar productDefinition. The scalar product of vectorsx = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn) isx · y = x1y1+ x2y2+ ··· + xnyn=nXk=1xkyk.Properties of scalar product:x · x ≥ 0, x ·x = 0 only if x = 0 (positivity)x · y = y · x (symmetry)(x + y) · z = x · z + y · z (distributive law)(rx) · y = r(x · y) (homogeneity)Relations between lengths and scalar products:kxk =√x · x|x · y| ≤ kxkkyk (Cauchy-Schwarz inequality)kx − yk2= kxk2+ kyk2− 2 x·yBy the Cauchy-Schwarz inequality, for any nonzerovectors x, y ∈ Rnwe havecos θ =x · ykxkkykfor some 0 ≤ θ ≤ π.θ is called the angle between the vectors x and y.The vectors x and y are said to be orthogonal(denoted x ⊥ y) if x · y = 0 (i.e., if θ = 90o).Problem. Find the angle θ between vectorsx = (2, −1) and y = (3, 1).x ·y = 5, kxk =√5, kyk =√10.cos θ =x ·ykxkkyk=5√5√10=1√2=⇒ θ = 45oProblem. Find the angle φ between vectorsv = (−2, 1, 3) and w = (4, 5, 1).v · w = 0 =⇒ v ⊥ w =⇒ φ =
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