Mathematics 44 -- Linear Algebra

Review True-False Questions -- Exam 3

April 21, 1999

• 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. In either case, the polynomial is a continuous function, so the Intermediate Value Theorem implies that there is some lambda value 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.
• E) Every invertible matrix is a product of elementary matrices E_ij(c), E_i(c), and P_ij for some i,j,c's -- TRUE. The echelon form of an invertible matrix is the identity matrix of the same size. Hence the Gauss-Jordan procedure yields Q₁ ... Q_k A = I. Solving for A shows that A = Q_k^-1 ... Q₁^-1 which is a product of elementary matrices.
• F) The eigenvalues of aE₁₁+bE₁₂ +bE₂₁+aE₂₂ are lambda = a+b and lambda = a-b -- TRUE, since T(1,1) = (a+b,a+b) and T(1,-1) = (a-b,b-a) .
• G) If the linear mapping T : R³ -> R³ has eigenvalues lambda = 1,2, and A₁,A₂ are linearly independent eigenvectors for lambda = 1, while B is an eigenvector for lambda = 2, then E = {A₁,A₂,B} is linearly independent -- TRUE. Consider a linear combination aA₁ + bA₂ + cB = 0. apply T to both sides. Then aA₁ + bA₂ + 2cB = 0. Subtract 2 times the previous equation from this one: aA₁ + bA₂ = 0. Since {A₁,A₂} is linearly independent, a = b = 0, but then c = 0 also.
• H) Every system of 3 linear equations with real coefficients in 5 variables has at least one solution in R⁵ -- FALSE. Write the system as Ax = b, or in augmented matrix form: [A|b]. This will fail if the vector b of right-hand sides is not in the linear span of the columns of the matrix A. In that case there is no solution. (The five columns of A must span at most a two-dimensional subspace of R³ for this to yield a counterexample.)