Review Notes for Linear Algebra True or False Last Updated: January 25, 2010 Chapter 3 [ Eigenvalues and Eigenvectors ] 31 If A is an n n matrix, then A can have at most n eigenvalues The characteristic equation det (A λi) = 0 is indeed a polynomial equation of degree n which has at most n roots (eigenvalues) 32 Any square matrix has at least one eigenvector Any polynomial equation has at least one root 33 If v is an eigenvector, then v 0 By definition, an eigenvector v satisfying Av = λv must be a nonzero vector 34 If λ is an eigenvalue, then λ 0 False It is permissible to have zero eigenvalue has eigenvalues 0, 1 35 If all eigenvalues of A are zero, then A = O has repeated eigenvalue 0 36 If all eigenvalues of A are 1, then A = I has repeated eigenvalue 1 37 If A I, then 1 is not an eigenvalue of A has eigenvalues 0, 1 38 If A O, then 0 is not an eigenvalue of A has repeated eigenvalue 0 39 If A 2 = O, then 0 may not be the only eigenvalue of A False The correct statement should be : If A 2 = O, then 0 is the only eigenvalue of A Let λ be the eigenvalue of A with eigenvector v Then Av = λv, A 2 v = A(λv) = λ 2 v Since A 2 = O and v 0, then λ = 0 is the only eigenvalue of A 310 If λ is an eigenvalue of A, then λ is also an eigenvalue of A t By det (A t λi) = det (A λi) t = det(a λi), then A and A t must have the same eigenvalues since they have exactly the same characteristic polynomial 311 If λ is an eigenvalue of A, then λ is also an eigenvalue of A 2 has eigenvalues ±1 But 1 is not an eigenvalue of A 1 0 2 = I 1
312 If λ is an eigenvalue of A and B, then λ is also an eigenvalue of A + B 1 0, B = Then 1 is an eigenvalue of A and B but 1 is not an eigenvalue of A + B = O 313 If λ is an eigenvalue of A and B, then λ is also an eigenvalue of AB False Take A, B in 312 Then 1 is not an eigenvalue of AB = [ ] 1 0 314 If λ is an eigenvalue of A and µ is an eigenvalue of B, then λµ is an eigenvalue of AB False Take A, B in 312 Then 1 is an eigenvalue of A and 1 is an eigenvalue of B, but 1 = ( 1)( 1) is not an eigenvalue of AB 315 If λ 0 is an eigenvalue of A, then λ 1 is an eigenvalue of A 1 (if exists) eigenvector Av = λv = A 1 Av = A 1 λv = λ 1 v = A 1 v = λ 1 is an eigenvalue of A 1 with the same 316 Suppose A is invertible If λ is an eigenvalue of A, then λ 1 is an eigenvalue of A 1 Av = λv = A 1 Av = A 1 λv = v = λ A 1 v Since A is invertible, by 329, 0 is never an eigenvalue Thus, λ 0 Hence, λ 1 v = A 1 v = λ 1 is an eigenvalue of A 1 with the same eigenvector 317 If u and v are eigenvectors of A, then u + v is an eigenvector of A [ ] [ 1 and u =, v = 0] [ ] 1 0 318 If u is an eigenvector of A and B, then u is an eigenvector of A + B If u is an eigenvector of A and B, then Au = λu and Bu = µu, for some λ, µ = (A + B)u = (λ + µ) u 319 If u is an eigenvector of A and B, then u is an eigenvector of AB (AB)u = A(Bu) = A(µu) = µ(au) = (µλ)u If u is an eigenvector of A and B, then Au = λu and Bu = µu, for some λ, µ = 320 If u and v are eigenvectors of A and B, then u + v is an eigenvector of A + B [ ] [ [ ] 1 0 B =, u = and v = But u + v = 0 can never be an eigenvector 1 1 1] 1 321 If v is an eigenvector of A 2, then v is an eigenvector of A [ ] [ 0 and v = 1 ] 2
322 If v is an eigenvector of A, then v is also an eigenvector of A 2 v is an eigenvector of A = Av = λv for some λ = A 2 v = A(Av) = A(λv) = λ(av) = λ 2 v 323 If v is an eigenvector of A, then v is also an eigenvector of A t [ ] [ 1 and v = 0] 324 If v is an eigenvector of A, then v is also an eigenvector of 2A (2A)v = 2Av = 2λv = (2λ)v 325 If v is an eigenvector of A, then 2v is again an eigenvector of A A(2v) = 2Av = 2λv = λ(2v) 326 If v 1, v 2 are eigenvectors of A corresponding to eigenvalues λ 1, λ 2, respectively Then v 1 + v 2 is an eigenvector of A corresponding to eigenvalue λ 1 + λ 2 has eigenvalues ±1, but 0 = 1 + 1 is not an eigenvalue of A 327 If v is an eigenvector of A and B, then v is an eigenvector of A 2 + 3AB (A 2 + 3AB)v = A(Av) + 3A(Bv) = A(λv) + 3A(µv) = λ(av) + 3µ(Av) = λ(λv) + 3µ(λv) = (λ 2 + 3λµ)v 328 If 0 is an eigenvalue of A, then A is not invertible Suppose 0 is an eigenvalue of A n n Then 0 is a root of the characteristic equation = det (A 0I n) = 0 = deta = 0 = A is not invertible 329 If A is invertible, then 0 is never an eigenvalue of A Suppose A is invertible Then det A 0 = λ = 0 does not satisfy the characteristic equation det (A λi) = 0 = 0 is not an eigenvalue of A 330 If all eigenvalues of A are nonzero, then A is invertible On the contrary, suppose A n n is not invertible Then deta = 0 = det(a λi n) = 0, λ = 0 = λ = 0 is a root of the characteristic equation = 0 is an eigenvalue of A 331 Distinct eigenvectors are linearly independent False By 325, v is an eigenvector of A = 2v is an eigenvector of A 3
332 Suppose u and v are eigenvectors of A with eigenvalues λ and µ If λ µ, then u and v are linearly independent We need to prove that c 1 u + c 2 v = 0 = c 1 = c 2 = 0 Proof: c 1 Au + c 2 Av = A0 = 0 Since u and v are eigenvectors, we have c 1 λu + c 2 µv = 0 Since λ and µ cannot be both zero, without the loss of generality, let us suppose λ 0 Solving the equation with c 1 λu + c 2 λv = 0 gives c 2 (λ µ)v = 0 Since v 0 and λ µ, then c 2 must be zero and hence c 1 must also be zero (u 0) 333 If A n n has n distinct eigenvalues, then there is a basis of eigenvectors A has n distinct eigenvalues = the corresponding n eigenvectors are linearly independent = the n eigenvectors form a basis of R n 334 All real symmetric matrices are diagonalizable For example, if A = 1 2 3 2 4 1, then the characteristic equation det (A λi) = 25 15λ + 10λ 2 λ 3 = 0 3 1 5 has no repeated root Hence all eigenvalues of A are distinct and A is diagonalizable 335 For any real matrix A, A t A is always diagonalizable For any real A, the matrix A t A is real symmetric: (A t A) t = A t (A t ) t = A t A It follows from 334 that A t A is diagonalizable 336 All diagonalizable matrices are symmetric False is diagonalizable but not symmetric 0 2 337 Any diagonal matrix is diagonalizable Diagonal matrix D always has a diagonalization: D = IDI 1, where I is an identity matrix 338 Any 1 1 matrix is diagonalizable A = [ a ] is always diagonal Then A = IAI 1 is a diagonalization of A 339 If A is invertible and diagonalizable, so is A 1 A is diagonalizable = invertible P and diagonal D such that P 1 AP = D = P 1 A 1 P = D 1, which is diagonal Also remark that, for example, if D = diag (λ 1, λ 2, λ 3 ) then D 1 = diag (1/λ 1, 1/λ 2, 1/λ 3 ) 340 If A is diagonalizable, then A t is also diagonalizable A is diagonalizable = invertible P and diagonal D such that P 1 AP = D = P t A t (P t ) 1 = D t = D 341 If A is diagonalizable, then A 3 is also diagonalizable A is diagonalizable = invertible P and diagonal D such that P 1 AP = D = P 1 A 3 P = D 3, which is diagonal Also remark that, for example, if D = diag (λ 1, λ 2, λ 3 ) then D 3 = diag(λ 3 1, λ3 2, λ3 3 ) 4
342 If A is diagonalizable, then A 2 is also diagonalizable Similar proof in 341 In fact, if A is diagonalizable, then A n is also diagonalizable, for n = ±1, ±2, 343 If A and B are diagonalizable, then AB is also diagonalizable [ ] 1 1 1 1, B = are diagonalizable But AB = is not 344 If A 3 is diagonalizable, then A is diagonalizable has repeated eigenvalue λ = 0 = A is not diagonalizable But A 3 = O is diagonal and hence diagonalizable 345 If A 2 is diagonalizable, then A is also diagonalizable False Take A in 344 346 If A n n has n distinct eigenvalues, then A is diagonalizable Distinct eigenvalues λ 1, λ 2,, λ n = corresponding eigenvectors v 1, v 2,, v n linearly independent = P = [ v 1 v 2 v n ] is invertible = A = PDP 1 is a diagonalization, where D = diag (λ 1, λ 2,, λ n) 347 If A n n is diagonalizable, then A must have n distinct eigenvalues 1 1 = 1 1 1 0 1 1 1 1 1 1 2 1 1 1 1 348 If A n n has fewer than n distinct eigenvalues, then A is not diagonalizable False A may have repeated eigenvalues but enough number of eigenvectors to form a diagonalization Take the 3 3 matrix A in 347 Then A has less than 3 distinct eigenvalues ( 1 is repeated) but it is still diagonalizable 349 If A and B have the same eigenvalues, then A is diagonalizable = B is also diagonalizable 1 1, B = have the same eigenvalues (repeated eigenvalue 1) But here only A is diagonalizable (A is diagonal and by 337) 350 If A is invertible and diagonalizable, and B is not diagonalizable, then AB is not diagonalizable 1 1, B = Then A is invertible (deta 0) and diagonalizable (A is diagonal and by 337), B is not diagonalizable (2 2 matrix with repeated eigenvalue), but AB = is diagonalizable (AB is upper-triangular with distinct diagonal entries) 351 If A is diagonalizable, then A is invertible is diagonal 352 If A is invertible, then A is diagonalizable has repeated eigenvalue 1 5