MATH 311-504Topics in Applied MathematicsLecture 2-13:Review for Test 2.Topics for Test 2Vector spaces and linear transformations (Williamson/Trotter3.1–3.4)• Vector spaces. Subspaces.• Linear mappings. Matrix transformations.• Span. Image and null-space.• Linear independence (especially in functional spaces).Basis, dimension, coordinates (Williamson/Trotter 3.5, 3.6C)• Basis of a vector space. Dimension.• Matrix of a linear transformation.• Change of coordinates.Eigenvalues and eigenvectors (Williamson/Trotter 3.6)• Eigenvalues, eigenvectors, eigenspaces.• Characteristic equation of a matrix.• Bases of eigenvectors, diagonalization.Sample problems for Test 2Problem 1 (20 pts.) Determine which of thefollowing subsets of R3are subspaces. Brieflyexplain.(i) The set S1of vectors (x, y, z) ∈ R3such thatxyz = 0.(ii) The set S2of vectors (x, y, z) ∈ R3such thatx + y + z = 0.(iii) The set S3of vectors (x, y, z) ∈ R3such thaty2+ z2= 0.(iv) The set S4of vectors (x, y, z) ∈ R3such thaty2− z2= 0.Sample problems for Test 2Problem 2 (20 pts.) Let M2,2(R) denote thespace of 2-by-2 matrices with real entries. Considera linear operator L : M2,2(R) → M2,2(R) given byLx yz w=1 23 4x yz w.Find the matrix of the operator L with respect tothe basisE1=1 00 0, E2=0 10 0, E3=0 01 0, E4=0 00 1.Sample problems for Test 2Problem 3 (30 pts.) Consider a linear operatorf : R3→ R3, f (x) = Ax, whereA =1 −1 −2−2 1 3−1 0 1.(i) Find a basis for the image of f .(ii) Find a basis for the null-space of f .Sample problems for Test 2Problem 4 (30 pts.) Let B =1 2 01 1 10 2 1.(i) Find all eigenvalues of the matrix B.(ii) For each eigenvalue of B, find an associatedeigenvector.(iii) Is there a basis for R3consisting ofeigenvectors of B? Explain.(iv) Find a diagonal matrix D and an invertiblematrix U such that B = UDU−1.(v) Find all eigenvalues of the matrix B2.Sample problems for Test 2Bonus Problem 5 (20 pts.) Solve the followingsystem of differential equations (find all solutions):dxdt= x + 2y,dydt= x + y + z,dzdt= 2y + z.Problem 1. Determine which of the followingsubsets of R3are subspaces. Briefly explain.A subset of R3is a subspace if it is closed under addition andscalar multiplication. Besides, the subset must not be empty.(i) The set S1of vectors (x, y, z) ∈ R3such thatxyz = 0.(0, 0, 0) ∈ S1=⇒ S1is not empty.xyz = 0 =⇒ (rx)(ry)(rz) = r3xyz = 0.That is, v = (x, y, z) ∈ S1=⇒ rv = (rx, ry , rz) ∈ S1.Hence S1is closed under scalar multiplication.However S1is not closed under addition.Counterexample: (1, 1, 0) + (0, 0, 1) = (1, 1, 1).Problem 1. Determine which of the followingsubsets of R3are subspaces. Briefly explain.A subset of R3is a subspace if it is closed under addition andscalar multiplication. Besides, the subset must not be empty.(ii) The set S2of vectors (x, y, z) ∈ R3such thatx + y + z = 0.(0, 0, 0) ∈ S2=⇒ S2is not empty.x + y + z = 0 =⇒ rx + ry + rz = r(x + y + z) = 0.Hence S2is closed under scalar multiplication.x + y + z = x′+ y′+ z′= 0 =⇒(x + x′) + (y + y′) + (z + z′) = (x + y + z) + (x′+ y′+ z′) = 0.That is, v = (x, y, z), v′= (x, y, z) ∈ S2=⇒ v + v′= (x + x′, y + y′, z + z′) ∈ S2.Hence S2is closed under addition.(iii) The set S3of vectors (x, y, z) ∈ R3such thaty2+ z2= 0.y2+ z2= 0 ⇐⇒ y = z = 0.S3is a nonempty set closed under addition and scalarmultiplication.(iv) The set S4of vectors (x, y, z) ∈ R3such thaty2− z2= 0.S4is a nonempty set closed under scalar multiplication.However S4is not closed under addition.Counterexample: (0, 1, 1) + (0, 1, −1) = (0, 2, 0).Problem 2. Let M2,2(R) denote the vector space of 2×2matrices with real entries. Consider a linear operatorL : M2,2(R) → M2,2(R) given byLx yz w=1 23 4x yz w.Find the matrix of the operator L with resp ect 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 matrix whose columns arecoordinates of the matrices L(E1), L(E2), L(E3), L(E4)with respect to the bas is E1, E2, E3, E4.L(E1) =1 23 41 00 0=1 03 0= 1E1+0E2+3E3+0E4,L(E2) =1 23 40 10 0=0 10 3= 0E1+1E2+0E3+3E4,L(E3) =1 23 40 01 0=2 04 0= 2E1+0E2+4E3+0E4,L(E4) =1 23 40 00 1=0 20 4= 0E1+2E2+0E3+4E4.It follows thatML=1 0 2 00 1 0 23 0 4 00 3 0 4.Thus the relationx1y1z1w1=1 23 4x yz wis equivalent to the relationx1y1z1w1=1 0 2 00 1 0 23 0 4 00 3 0 4xyzw.Problem 3. Consider a linear operator f : R3→ R3,f (x) = Ax, where A =1 −1 −2−2 1 3−1 0 1.(i) Find a basis for the image of f .The image of f is spanned by columns of the matrix A:v1= (1, −2, −1), v2= (−1, 1, 0), v3= (−2, 3, 1).det A =1 −1 −2−2 1 3−1 0 1= −1−1 −21 3+ 11 −1−2 1= 0.Hence v1, v2, v3are linearly dependent.It is easy to observe that v2= v1+ v3.It follows that Span(v1, v2, v3) = Span(v1, v3).Since the vectors v1and v3are linearly independent, they forma basis for the image of f .Problem 3. Consider a linear operator f : R3→ R3,f (x) = Ax, where A =1 −1 −2−2 1 3−1 0 1.(ii) Find a basis for the null-space of f .The null-space of f is the set of solutions of the vectorequation Ax = 0. To solve the equation, we convert thematrix A to reduced row echelon form:1 −1 −2−2 1 3−1 0 1→1 −1 −20 −1 −10 −1 −1→1 −1 −20 −1 −10 0 0→1 −1 −20 1 10 0 0→1 0 −10 1 10 0 0→x − z = 0,y + z = 0.General solution: (x, y , z) = (t, −t, t) = t(1, −1, 1), t ∈ R.Hence the null-space is a line and (1, −1, 1) is its basis.Problem 4. Let B =1 2 01 1 10 2 1.(i) Find all eigenvalues of the matrix B.The eigenvalues of B are roots of the characteristic equationdet(B − λI ) = 0. We obtain thatdet(B − λI ) =1 − λ 2 01 1 − λ 10 2 1 − λ= (1 − λ)3− 2(1 − λ) − 2(1 − λ) = (1 − λ)(1 − λ)2− 4= (1 − λ)(1 − λ) − 2(1 − λ) + 2= −(λ − 1)(λ + 1)(λ − 3).Hence the matrix B has three eigenvalues: