MATH 304Linear AlgebraLecture 34:Review for Test 2.Topics for Test 2Coordinates and linear transformations (Leon 3.5, 4.1–4.3)• Coordinates relative to a basis• Change of basis, transition matrix• Matrix transformations• Matrix of a linear ma ppingOrthogonality (Leon 5.1–5.6)• Inner products and norms• Orthogonal complement, orthogonal projection• Least squares problems• The Gram-Schmidt orthogonalization processEigenvalues and eigenvectors (Leon 6.1, 6.3)• Eigenvalues, eigenvectors, eigenspaces• Characteristic polynomial• DiagonalizationSample problems for Test 2Problem 1 (15 pts.) Let M2,2(R) denote the v ector spaceof 2 × 2 matrices with real entries. Consider a linear operatorL : M2,2(R) → M2,2(R) given byLx yz w=x yz w1 23 4.Find the matrix of the operator L with respect to the basisE1=1 00 0, E2=0 10 0, E3=0 01 0, E4=0 00 1.Problem 2 (20 pts.) Find a linear polynomial which is thebest least squares fit to the following data:x−2 −1 0 1 2f (x) −3 −2 1 2 5Problem 3 (25 pts.) Let V be a subspace of R4spannedby the vectors x1= (1, 1, 1, 1) and x2= (1, 0, 3, 0).(i) Find an orthonormal basis f or V .(ii) Find an orthonormal basis for the orthogonal complementV⊥.Problem 4 (30 pts.) Let A =1 2 01 1 10 2 1.(i) Find all eigenvalues of the matrix A.(ii) For each eigenvalue of A, find an associated eigenvector.(iii) Is the matrix A diagonalizable? Explain.(iv) Find all eigenvalues of the matrix A2.Bonus Problem 5 (15 pts.) Let L : V → W be a linearmapping of a finite-dimensional vector space V to a vectorspace W . Show thatdim Range(L) + dim ker(L) = dim V .Problem 1. Let M2,2(R) denote the v ector space o f 2×2matrices with real entries. Consider a linear operatorL : M2,2(R) → M2,2(R) given byLx yz w=x yz w1 23 4.Find the matrix of the operator L with respect to the basisE1=1 00 0, E2=0 10 0, E3=0 01 0, E4=0 00 1.Let MLdenote the desired matrix.By definition, MLis a 4×4 ma trix whose columns arecoordinates of the matrices L(E1), L(E2), L(E3), L(E4)with respect to the basis E1, E2, E3, E4.L(E1) =1 00 01 23 4=1 20 0= 1E1+ 2 E2+ 0 E3+ 0 E4,L(E2) =0 10 01 23 4=3 40 0= 3E1+ 4 E2+ 0 E3+ 0 E4,L(E3) =0 01 01 23 4=0 01 2= 0E1+0E2+1E3+2E4,L(E4) =0 00 11 23 4=0 03 4= 0E1+0E2+3E3+4E4.It follows thatML=1 3 0 02 4 0 00 0 1 30 0 2 4.Thus the relationx1y1z1w1=x yz w1 23 4is equivalent to the relationx1y1z1w1=1 3 0 02 4 0 00 0 1 30 0 2 4xyzw.Problem 2. Find a linear polynomial which is the best leastsquares fit to the following data:x−2 −1 0 1 2f (x) −3 −2 1 2 5We are looking for a function f (x) = c1+ c2x, where c1, c2are unknown coefficients. The data of the problem give riseto an overdetermined system of linear equations in variables c1and c2:c1− 2c2= −3,c1− c2= −2,c1= 1,c1+ c2= 2,c1+ 2c2= 5.This system is inconsistent.We can represent the system as a matrix equation Ac = y,whereA =1 −21 −11 01 11 2, c =c1c2, y =−3−2125.The least squares solution c of the above system is a solutionof the normal system ATAc = ATy:1 1 1 1 1−2 −1 0 1 21 −21 −11 01 11 2c1c2=1 1 1 1 1−2 −1 0 1 2−3−2125⇐⇒5 00 10c1c2=320⇐⇒c1= 3/5c2= 2Thus the function f (x) =35+ 2x is the best least squares fitto the above data among linear polynomials.Problem 3. Let V be a subspace of R4spanned by thevectors x1= (1, 1, 1, 1) and x2= (1, 0, 3, 0).(i) Find an orthonormal basis f or V .First we apply the Gram-Schmidt orthogonalization process tovectors x1, x2and obtain an orthogonal basis v1, v2for thesubspace V :v1= x1= (1, 1, 1, 1 ),v2= x2−x2· v1v1· v1v1= (1, 0, 3, 0)−44(1, 1, 1, 1) = (0, −1, 2, −1).Then we normalize vectors v1, v2to obtain an orthonormalbasis w1, w2for V :kv1k = 2 =⇒ w1=v1kv1k=12(1, 1, 1, 1)kv2k =√6 =⇒ w2=v2kv2k=1√6(0, −1, 2, −1)Problem 3. Let V be a subspace of R4spanned by thevectors x1= (1, 1, 1, 1) and x2= (1, 0, 3, 0).(ii) Find an orthonormal basis for the orthogonal complementV⊥.Since the subspace V is spanned by vectors (1, 1, 1, 1) and(1, 0, 3, 0), it is the row space of the matrixA =1 1 1 11 0 3 0.Then the orthogonal complement V⊥is the nullspace of A.To find the nullspace, we convert the matrix A to reduced rowechelon form:1 1 1 11 0 3 0→1 0 3 01 1 1 1→1 0 3 001 −2 1.Hence a vector (x1, x2, x3, x4) ∈ R4belongs to V⊥if and onlyif1 0 3 00 1 −2 1x1x2x3x4=00⇐⇒x1+ 3x3= 0x2− 2x3+ x4= 0⇐⇒x1= −3x3x2= 2x3− x4The general solution of the system is (x1, x2, x3, x4) == (−3t, 2t − s, t, s) = t(−3, 2, 1, 0) + s(0, −1, 0, 1), wheret, s ∈ R.It follows that V⊥is spanned by vectors x3= (0, −1, 0, 1)and x4= (−3, 2, 1, 0 ) .The vectors x3= (0, −1, 0, 1 ) and x4= (−3, 2, 1, 0 ) form abasis for the subspace V⊥.It remains to orthogonalize and normalize this basis:v3= x3= (0, −1, 0, 1 ) ,v4= x4−x4· v3v3· v3v3= (−3, 2, 1, 0 ) −−22(0, −1, 0, 1)= (−3, 1, 1 , 1),kv3k =√2 =⇒ w3=v3kv3k=1√2(0, −1, 0, 1),kv4k =√12 = 2√3 =⇒ w4=v4kv4k=12√3(−3, 1, 1, 1).Thus the vectors w3=1√2(0, −1, 0, 1) andw4=12√3(−3, 1, 1, 1) form an orthonormal basis for V⊥.Problem 3. Let V be a subspace of R4spanned by thevectors x1= (1, 1, 1, 1 ) a nd x2= (1, 0, 3, 0).(i) Find an orthonormal basis f or V .(ii) Find an orthonormal basis for the orthogonal complementV⊥.Alternative solution: First we extend the set x1, x2to a basisx1, x2, x3, x4for R4. Then we orthogonalize and normalizethe latter. This yields an orthonormal basis w1, w2, w3, w4for R4.By construction, w1, w2is an orthonormal basis for V .It follows that w3, w4is an orthonormal basis for V⊥.The set x1= (1, 1, 1, 1), x2= (1, 0, 3, 0 ) can be extended toa basis for R4by adding two vectors from the standard basis.For example, we can add vectors e3= (0, 0, 1, 0 ) ande4= (0, 0, 0, 1). To show that x1, x2, e3, e4is indeed a basisfor R4, we check that the matrix whose rows are these vectorsis nonsingular:1 1 1 11 0 3 00 0
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