Download THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences ... and more Lecture notes Linear Algebra in PDF only on Docsity! THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences SECOND TEST - SPRING SESSION 2005 110.201 - LINEAR ALGEBRA. Examiner: Professor C. Consani Duration: 50 minutes, April 27, 2005 No calculators allowed. Total Marks = 100 1. [25 marks] In R3, find the point P on the plane described by the equation x + y − z = 0 which is closest to b = 2 1 0 . Sol. Every point on the plane described by the equation x + y − z = 0 is a solution to [ 1 1 −1 ] x y z = 0. The special solutions −1 1 0 , 1 0 1 are a basis for the 2-dimensional plane in R3. The least squares solution x to the system −1 1 1 0 0 1 x = 2 1 0 determines the point P that is closest to b. Let A = −1 1 1 0 0 1 . In particular we get AT Ax = AT b that is[ 2 −1 −1 2 ] x = [−1 2 ] i.e. x = [ 0 1 ] . Hence, A [ 0 1 ] = 1 0 1 and P = (1, 0, 1). 2. [25 marks] Give an orthonormal basis for the image of the linear transformation described by the matrix 1 3 8 1 3 0 1 −1 0 1 −1 0 . Sol. Use the Gram-Schmidt process: v1 = 1 1 1 1 ; u1 = 1 2 1 1 1 1 v2 = 3 3 −1 −1 − (u1 · 3 3 −1 −1 )u1 = 2 2 −2 −2 ; u2 = 1 2 1 1 −1 −1 v3 = 8 0 0 0 − (u1 · 8 0 0 0 )u1 − (u2 · 8 0 0 0 )u2 = 4 −4 0 0 ; u3 = 1√ 2 1 −1 0 0 . 3. Consider the matrix A = 0 0 0 1 0 0 2 2 0 3 3 3 4 4 4 4 a) [10 marks] Find det(A). Sol. We can obtain an upper-traingular matrix via two row-exchanges. Ex- change rows 1 and 4; exchange rows 2 and 3. Two row exchanges: determinant does not change sign. The determinant of the upper-triangular matrix is: 1 · 2 · 3 · 4 = 24. b) [10 marks] Find det(1 2 A). Sol. det(1 2 A) = (1 2 )4 det(A) = 3 2 c) [5 marks] Is A diagonalizable? Why? Sol. The 4 eigenvalues of A are distinct, hence A is diagonalizable. 4. Suppose the following information is known about a matrix A A 1 2 1 = 6 1 2 1 , A 1 −1 1 = 3 1 −1 1 , A 2 −1 0 = 3 1 −1 1 . a) [10 marks] Find the eigenvalues of A. Sol. The first two results show that 6 and 3 are eigenvalues of A. The last two results show that there are (at least) two different solutions to the system Ax = 3 −3 3 . In other words, A has non-trivial nullspace, which means Ax = 0 0 0 = λ3x
