MATH 311Topics in Applied MathematicsLecture 16:Diagonalization.Euclidean structure in Rn.DiagonalizationLet L be a linear operator on a finite-dimensional vector spaceV . Then the following conditions are equivalent:• the matrix of L with respect to some basis is diagonal;• there exists a basis for V formed by eigenvectors of L.The operator L is diagonalizable if it satisfies theseconditions.Let A be an n×n matrix. Then the following conditions areequivalent:• A is the matrix of a diagonalizable operator;• A is similar to a diagonal matrix, i.e., it is represented asA = UBU−1, where the matrix B is diagonal;• there exists a basis for Rnformed by eigenvectors of A.The matrix A is diagonalizable if it satisfies these conditions.Otherwise A is called defective.Example. A =2 11 2.• The matrix A has two eigenvalues: 1 and 3.• The eigenspace of A associated with theeigenvalue 1 is the line spanned by v1= (−1, 1).• The eigenspace of A associated with theeigenvalue 3 is the line spanned by v2= (1, 1).• Eigenvectors v1and v2form a basis for R2.Thus the matrix A is diagonalizable. Namely,A = UBU−1, whereB =1 00 3, U =−1 11 1.Example. A =1 1 −11 1 10 0 2.• The matrix A has two eigenvalues: 0 and 2.• The eigenspace corresponding to 0 is spanned byv1= (−1, 1, 0).• The eigenspace corresponding to 2 is spanned byv2= (1, 1, 0) and v3= (−1, 0, 1).• Eigenvectors v1, v2, v3form a basis for R3.Thus the matrix A is diagonalizable. Namely,A = UBU−1, whereB =0 0 00 2 00 0 2, U =−1 1 −11 1 00 0 1.Problem. Diagonalize the matrix A =4 30 1.We need to find a diagonal matrix B and aninvertible matrix U such that A = UBU−1.Suppose that v1= (x1, y1), v2= (x2, y2) is a basisfor R2formed by eigenvectors of A, i.e., Avi= λivifor some λi∈ R. Then we can takeB =λ100 λ2, U =x1x2y1y2.Note that U is the transition matrix from the basisv1, v2to the standard basis.Problem. Diagonalize the matrix A =4 30 1.Characteristic equation of A:4 −λ 30 1 − λ= 0.(4 − λ)(1 − λ) = 0 =⇒ λ1= 4, λ2= 1.Associated eigenvectors: v1= (1, 0), v2= (−1, 1).Thus A = UBU−1, whereB =4 00 1, U =1 −10 1.Problem. Let A =4 30 1. Find A5.We know that A = UBU−1, whereB =4 00 1, U =1 −10 1.Then A5= UBU−1UBU−1UBU−1UBU−1UBU−1= UB5U−1=1 −10 11024 00 11 10 1=1024 −10 11 10 1=1024 10230 1.Problem. Let A =4 30 1. Find a matrix Csuch that C2= A.We know that A = UBU−1, whereB =4 00 1, U =1 −10 1.Suppose that D2= B for some matrix D. Let C = UDU−1.Then C2= UDU−1UDU−1= UD2U−1= UBU−1= A.We can take D =√4 00√1=2 00 1.Then C =1 −10 12 00 11 10 1=2 10 1.System of linear ODEsProblem. Solve a system(dxdt= 4x + 3y,dydt= y.The system can be rewritten in vector form:dvdt= Av, where A =4 30 1, v =xy.We know that A = UBU−1, whereB =4 00 1, U =1 −10 1.Let w =w1w2be coordinates of the vector v relative to thebasis v1= (1, 0), v2= (−1, 1) of eigenvectors of A. Thenv = Uw =⇒ w = U−1v.It follows thatdwdt=ddt(U−1v) = U−1dvdt= U−1Av = U−1AUw.Thusdwdt= Bw ⇐⇒(dw1dt= 4w1,dw2dt= w2.The general solution: w1(t) = c1e4t, w2(t) = c2et,where c1, c2are arbitrary constants. Thenx(t)y(t)= Uw(t) =1 −10 1c1e4tc2et=c1e4t−c2etc2et.There are two obstructions to diagonalization.They are illustrated by the following examples.Example 1. A =1 10 1.det(A − λI ) = (λ − 1)2. Hence λ = 1 is the onlyeigenvalue. The associated eigenspace is the linet(1, 0).Example 2. A =0 −11 0.det(A − λI ) = λ2+ 1.=⇒ no real eigenvalues or eigenvectors(However there are complex eigenvalues/eigenvectors.)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 r v 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 coordinate 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, . . . ,