Mathematics 244 -- Linear Algebra

Review True-False Questions -- Exam 3

April 29, 2004

• A) If A is a 3 x 3 matrix whose characteristic polynomial has a double root, then A is not diagonalizable -- FALSE. A diagonal matrix with 2 1's and a 2 on the diagonal is a counterexample -- the characteristic polynomial is (1 - lambda)²(2-lambda)
• B) If T: R² -> R² and there exists a non-zero vector x that satisfies T²(x) = 4x, then T(x) = + 2x or -2x -- FALSE. For example, let T(1,0) = (0,1) and T(0,1) = (4,0). Then T²(1,0) = 4(1,0), but T(1,0) = (0,1) is neither (2,0) nor (-2,0). (But it is true that T has some eigenvector for lambda = + or - 2. Say T(x) = y, so T(y) = 4x. Consider the vector z = 2x + y. Then T(z) = 2T(x) + T(y) = 2y + 4x = 2(2x + y) = 2z. If z = 0, then a similar argument shows w = -2x + y is an eigenvector for lambda = -2.)
• C) Every 3 x 3 matrix has at least one eigenvalue lambda in R -- TRUE. The characteristic polynomial p(lambda) is a polynomial of degree three with real coefficients. Recall the shape of the graph of a cubic polynomial. The coefficient of lambda³ is negative, so the limit as lambda -> -infinity is +infinity and the limit as lambda -> +infinity is -infinity. The polynomial is a continuous function, so the Intermediate Value Theorem implies that there is some lambda where the polynomial takes the value zero. Hence it has at least one (and no more than three) real root(s).
• D) If lambda is an eigenvalue of (A-sI)^-1, where s in R, then s + 1/lambda is an eigenvalue of A --TRUE. Let x be an eigenvector for (A-sI)^-1 with eigenvalue lambda. then (A-sI)^-1x = lambda x. Multiplying both sides by (A-sI), we get x = lambda Ax - lambda sx. Rearranging this, we get Ax = (s + 1/lambda)x. Thus x is an eigenvector for A with eigenvalue s + 1/lambda.
(This fact is used in Numerical Linear Algebra to derive a good method for computing eigenvalues for large matrices, called the Power Method.)
• E) The eigenvalues of aE₁₁+bE₁₂ +bE₂₁+aE₂₂ are lambda = a+b and lambda = a-b -- TRUE. The characteristic polynomial is lambda² - 2a lambda + (a² - b²), which factors as (lambda - (a+b))(lambda - (a - b)). Corresponding eigenvectors are T(1,1) = (a+b,a+b) = (a+b)(1,1) and T(1,-1) = (a-b,b-a) = (a-b)(1,-1).
• F) If the linear mapping T : R³ -> R³ has eigenvalues lambda = 1,2, and x₁,x₂ are linearly independent eigenvectors for lambda = 1, while y is an eigenvector for lambda = 2, then E = {x₁,x₂,y} is linearly independent -- TRUE. Consider a linear combination a x₁ + b x₂ + cy = 0. Apply T to both sides. Then a x₁ + b x₂ + 2c y = 0. Subtract 2 times the previous equation from this one: a x₁ + b x₂ = 0. Since {x₁,x₂} is linearly independent, a = b = 0, but then c = 0 also. (This is a special case of the third ``Proof to Know'' on the review sheet.)
• G) If A is an n x n matrix with linearly independent rows, then det(A) = 0 -- FALSE. In fact the determinant must be nonzero, because A is invertible in this case. To see this, recall that we know det(A) = det(A^t). The columns of A^t are linearly independent which says that they form a basis of Rⁿ. So the Dimension Theorem says the dimension of the image of A^t is n and the dimension of the kernel of A^t is zero. Hence A^t is invertible, so det(A^t) = det(A) is nonzero. (Note: It is a general fact for all matrices (not necessarily square) that the dimension of the span of the rows is the same as the dimension of the span of the columns. This can be seen by reducing the matrix to echelon form.)
• H) Let V be a vector space of dimension n and T: V -> V be a linear mapping. If [T]_a^a is the identity matrix for some basis a, then T is the identity mapping -- TRUE. Let a = {x₁, ... , x_n}. Since the matrix of T with respect to a is the identity matrix, T(x_i) = x_i for all i = 1, ... , n. Since T is linear, this implies T(x) = x for all x in V. So T is the identity mapping.