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Stony Brook University Mathematics Department Julia Viro Introduction to Linear Algebra MAT 211 March 6, 2008 Solutions for Midterm I Problem 1. Solve the system using the Gauss-Jordan elimination and verify your answer + x4 = 0 x1 + 3x2 x2 + x3 =−1 − x2 − x3 + x4 = 3. Solution. Elementary row transformations of the augmented matrix of the system give rise to the reduced row-echelon form: 1 3 0 1 1 0 −3 0 1 3 0 1 0 0 1 0 0 1 1 1 0 −1 ∼ 0 1 1 0 −1 1 0 −1 . ∼ R3+R2 R1−3R2−R3 3 2 2 0 −1 −1 1 0 0 0 1 0 0 0 1 Obviously, the rank of the matrix is 3. The number of free variables is 4 − 3 = 1 (the number of unknowns minus the rank). We write down the solution starting from the back: x4 = 2 (that is what the last row in the rref says), x3 = t (choose x3 as a free variable), x2 = −1 − t (from the second row in the rref), x1 = 1 + 3t (from the first row in the rref). So the solution is (x1 , x2 , x3 , x4 ) = (1 + 3t, −1 − t, t, 2) = (1, −1, 0, 2) + t(3, −1, 1, 0), where t is an arbitrary real number. Geometrically, the solution is a line in R4 . To verify the solution, we substitute it into the three equations of the system: 1 + 3t + 3(−1 − t) + 2 = 0 −1 − t + t = −1 . −(−1 − t) − t + 2 =3 Since all the equations are satisfied, our solution is correct. Answer: (x1 , x2 , x3 , x4 ) = (1 + 3t, −1 − t, t, 2), t ∈ R. 2 Problem 2. Let T : R2 → R2 be an orthogonal projection onto the line x − 2y = 0 followed by a counterclockwise rotation by 45◦ . Find a matrix of T (with respect to the standard basis). Describe geometrically and show on a picture the kernel and the image of T . Is T invertible? Explain! Solution. Denote by P the orthogonal projection onto the line x − 2y = 0 and by R the counterclockwise rotation by 45◦ . Then T = R ◦ P . Let A and B be standard matrices of P and R respectively. Then the standard matrix of T is BA. Let us evaluate matrices A and B. Matrix A consists of two columns representing the coordinates of the images of the standard basis vectors e1 , e2 under the transformation P : | | A = P e1 P e2 . | | To evaluate the images we use a formula defining the orthogonal projection P : x·u Px = u, kuk2 where x is an arbitrary vector in R2 , u is a vector along the line of projection, kuk is its length and · is a dot product. Take u = (2, 1). (Note that one can take any vector u√= (x, y) whose √ coordinate satisfy the equation of the line x − 2y = 0.) Its length is kuk = 22 + 12 = 5. Calculate the images of e1 = (1, 0) and e2 = (0, 1) under projection P : (1, 0) · (2, 1) 2 4 2 P e1 = (2, 1) = (2, 1) = , , 5 5 5 5 2 1 1 (0, 1) · (2, 1) (2, 1) = (2, 1) = , . P e2 = 5 5 5 5 It gives us the matrix A: 1 4 2 4/5 2/5 A= = . 2/5 1/5 5 2 1 The standard matrix of a counterclockwise rotation by 45◦ is √ √ √ 2 1 −1 cos 45◦ − sin 45◦ 2/2 − 2/2 √ B= = √ = . sin 45◦ cos 45◦ 2/2 2/2 2 1 1 The standard matrix of T is √ √ 2 1 −1 1 4 2 2 2 1 BA = = . 2 1 1 5 2 1 10 6 3 The kernel of T is a subspace which is annihilated by T . This is a line passing through the origin which is orthogonal to the line x − 2y = 0. The equation of this line is 2x + y = 0. Note that Ker T is spanned by vector (1, −2) which is annihilated by T . The image of T is the line x − 2y = 0 rotated counterclockwise by 45◦ around the origin. The 1 equation of this line is 3x − y = 0. Note that Im T is spanned by a column vector of the 3 matrix of T . 3 Transformation T is not invertible. It can be explained in many different ways. For example, Ker T 6= 0 or Im T 6= R2 or the determinant of the matrix of T is 0. The picture makes all calculations crystal clear: Ker T Im T 2y= x− 3x− 0 y= Answer: 2x+ y=0 45 √ 2 2 1 the standard matrix of T is , 10 6 3 Ker T = span {(1, −2)}, Im T = span {(1, 3)}, T is not invertible. 0 4 Problem 3. A secret agent has got an encoded message 7, 12, −12, 16, 24, −24 representing the date of a secret operation. He knows that the coding matrix is 1 0 2 0 1 3 , −2 0 −3 but nevertheless cannot decode since he is not good in Linear Algebra. Help him to decode the secret date! Solution. Oh, boy! First, invert the matix: 1 0 2 1 0 0 1 0 2 1 0 0 1 0 0 −3 0 −2 0 1 3 0 1 0 ∼ 0 1 3 0 1 0 ∼ 0 1 0 −6 1 −3 R3+2R1 R1−2R3 −2 0 −3 0 0 1 0 0 1 2 0 1 0 0 1 2 0 1 R2−3R3 Second, multiply the inverse matrix by the two vectors (7, 12, −12) and (16, 24, −24) from the encoded message: 3 7 −3 0 2 −6 1 −3 12 = 6 , 2 −12 2 0 1 0 16 −3 0 2 −6 1 −3 24 = 0 . 8 −24 2 0 1 Third, read the secret date: 3, 6, 2, 0, 0, 8 or 3/6/2008. (The secret operation is Midterm I : -) Answer: 3/6/2008 5 Problem 4. Let V be a subspace of R4 given by V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0}. Find a basis in V . Show that the vector v = (0, 1, −1, 2) belongs to V and find the coordinates of v with respect to the basis you have found. Solution. The subspace V consists of all the solutions of the homogeneous system x1 + 2x2 − x4 x2 + x3 = 0 = 0. En elementary row transformation of the augmented matrix gives rise to the reduced rowechelon form: 1 2 0 −1 0 0 1 1 0 0 ∼ R1−2R2 1 0 −2 −1 0 0 1 1 0 0 . The number of free variables in the solution is 4 − 2 = 2 (the number of unknowns minus the rank of the matrix). We put x4 = t and x3 = s ( t, s are arbitrary real numbers) and get x2 = −s and x1 = t + 2s. It gives us the solution in the form (x1 , x2 , x3 , x4 ) = (t + 2s, −s, s, t) = t(1, 0, 0, 1) + s(2, −1, 1, 0), t, s ∈ R. Now we see that the subspace V is spanned by two vectors, (1, 0, 0, 1) and (2, −1, 1, 0). They comprise a basis of V . Note that a basis is not unique and other answers are possible. One can get a basis of V in a short cut: V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0} = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 = −2x2 + x4 , x2 = −x3 } = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 = 2x3 + x4 , x2 = −x3 } = {(2x3 + x4 , −x3 , x3 , x4 ) | x3 , x4 ∈ R} = {x3 (2, −1, 1, 0) + x4 (1, 0, 0, 1) | x3 , x4 ∈ R} = span{(2, −1, 1, 0), (1, 0, 0, 1)}. Now let us check that the vector v = (0, 1, −1, 2) belongs to V . Its components x1 = 0, x2 = 1, x3 = −1, x4 = 2 satisfy the two equations defining V : 0 + 2 · 1 − 2 = 0 and 1 + (−1) = 0. In order to find the coordinates of v with respect to the basis B = {(1, 0, 0, 1), (2, −1, 1, 0)}, we organize a linear combination v = (0, 1, −1, 2) = a(1, 0, 0, 1) + b(2, −1, 1, 0) and find the coefficients a and b: a = 2, b = −1. Answer: A basis of V is B = {(1, 0, 0, 1), (2, −1, 1, 0)}. Coordinates of v = (0, 1, −1, 2) with respect to B are (2, −1). 6 Problem 5. A linear transformation T : R3 → R4 is defined by T (x, y, z) = (y + 2z, −y + z, 3y + z, 4y + 2z). a) Find the matrix of T with respect to the standard bases. b) Find a basis in the kernel of T and a basis in the image of T . c) Find the dimensions of the kernel and the image. d) Find the rank of T . e) Verify the Kernel-Image theorem for T . Solution. The matrix of T with respect to the standard bases in R3 and R4 is 0 1 2 0 −1 1 A= 0 3 1 0 4 2 4×3 We perfom elementary row transformations to get the reduced row-echelon form of A: 0 1 0 0 1 2 0 1 2 0 −1 1 0 0 3 0 0 1 A= 0 3 1 ∼ 0 0 −5 ∼ 0 0 0 = rref(A). 0 0 0 0 0 −6 0 4 2 Obviously, the rank of A is 2. It is the dimension of the image of T . The image of T is generated by the second and the third columns of A, since the leading ones in the rref(A) stay in the second and the third columns: Im T = span{(1, −1, 3, 4), (2, 1, 1, 2)}. Since the spanning vectors are linearly independent they comprise a basis of T . The Kernel-Image Theorem says that dim Ker T + dim Im T = dim R3 or dim Ker T + 2 = 3. So dim Ker T = 1. A basis of Ker T can be found by solving a homogenous linear system with coefficient matrix A. It is easy to read the solution from the rref(A): z = 0, y = 0, x = t, where t is an arbitrary real number. Hence Ker T = {(t, 0, 0) | t ∈ R} = span{(1, 0, 0)}. Answer: 0 1 2 0 −1 1 The standard matrix of T is 0 3 1 , 0 4 2 a basis of Ker T is {(1, 0, 0)}, a basis of Im T is {(1, −1, 3, 4), (2, 1, 1, 2)}, dim Ker T = 1, dim Im T = rk T = 2.