True False Practice

Math 217: True False Practice 1. A square matrix is invertible if and only if zero is not an eigenvalue. Solution note: True. Zero is an eigenvalue means that there is a non-zero element in the kernel. For a square matrix, being invertible is the same as having kernel zero. 2. If A and B are 2 × 2 matrices, both with eigenvalue 5, then AB also has eigenvalue 5. Solution note: are 25. False. This is silly. Let A = B = 5I2 . Then the eigenvalues of AB 3. If A and B are 2 × 2 matrices, both with eigenvalue 5, then A + B also has eigenvalue 5. Solution note: False. This is silly. Let A = B = 5I2 . Then the eigenvalues of A + B are 10. 4. A square matrix has determinant zero if and only if zero is an eigenvalue. Solution note: True. Both conditions are the same as the kernel being non-zero. T 5. If B is the B-matrix of some linear transformation V → V . Then for all ~v ∈ V , we have B[~v ]B = [T (~v )]B . Solution note: True. This is the definition of B-matrix.  1 2 6. Suppose 0 2 0 0 B = (f1 , f2 , f3 ).  3 T 0 is the matrix of a transformation V → V with respect to some basis 1 Then f1 is an eigenvector. Solution note: True. It has eigenvalue 1. The first column of the B-matrix is telling us that T (f1 ) = f1 .  1 2  7. Suppose 0 2 0 0 B = (f1 , f2 , f3 ).  3 T 0 is the matrix of a transformation V → V with respect to some basis 1 Then T (f1 + f2 + f3 ) is 6f1 + 2f2 + f3 .   1 Solution note: TRUE! The B-coordinates of f1 + f2 + f3 are 1. To get the B1   1 2 3 coordinates of T (f1 + f2 + f3 ) we just multiply by the matrix [T ]B = 0 2 0 to 0 0 1   6 get 2. This represents the vector 6f1 + 2f2 + f3 . 1       1 0 0 0 0 1 8. The matrices , , and form a basis for the space of symmetric 2 × 2 0 0 0 1 1 0 matrices. Solution note: TRUE. They are clearly all in the space of symmetric matrices and are linearly independent. But the space of symmetric matrices has dimension less than 4, since not every matrix is symmetric. So it must have dimension 3, in which case these are a basis. 9. The only rotation R2 → R2 which has a real eigenvalue are rotations that induce the identity transformation (so through ±2π, ±4π, etc). Solution note: FALSE! Rotation through π has eigenvalue -1.   10. If the change of basis matrix SA→B = ~e4 ~e3 ~e2 ~e1 , then the elements of A are the same as the element of B, but in a different order. Solution note: True. The matrix tells us that the first element of A is the fourth element of B, the second element of basis A is the third element of B, the third element of basis A is the second element of B, and the the fourth element of basis A is the first element of B. 11. The map assigning hA, Bi to trace(AB T ) is an inner product on the space of all R2×2 matrices. Solution note: TRUE. It satisifies the four axioms in 5.5. 12. An orthogonal matrix must have at least one real eigenvalue. Solution note: False! Rotation through 90 degrees is orthogonal but has no real eigenvalues! 13. Both hA, Bii = trace AT B and hA, Bi = trace AB T define an inner product on R2×2 . Solution note: True! Both satisfy the axioms of 5.5 14. If A is a 3 × 4 matrix, then the matrix AT A is similar to a diagonal matrix with three or less non-zero entries. Solution note: True! The matrix AT A is symmetric, so by the spectral theorem, it is similar to a diagonal matrix. But also, its rank is at most 3 since we had a homework exercise in which we checked that the rank of a matrix can not go up when we multiply by any other matrix, so rank AT A can’t be more than rank A which is at most 3 since A is 3 × 4. So the 4 × 4 matrix AT A has rank at most 3 which means it is not invertible. This means zero is an eigenvalue. Since the eigenvalues are the elements on the diagonal of this diagonal matrix, this means there is a zero on the diagonal, so at most 3 non-zero entries. 15. If A is similar to both D1 and D2 , where D1 and D2 are diagonal, then D1 = D2 . Solution note: False! The elements on the diagonal are the  eigenvalues,   but they 1 0 2 0 could be arranged in different orders. For example, and are similar, 0 2 0 1   0 1 taking S = . 1 0 16. Let u and v be any two orthonormal vectors in an inner product space. Then ||u − v|| = √ 2. Solution note: True. ||u−v|| is the square root of (u−v)·(u−v) = u·u−2u·v+v·v = 2. 17. If hx, yi = −hy, xi in some inner product space, then x is orthogonal to y. Solution note: True! We know hx, yi = hy, xi by the symmetric property of inner products, so the hypothesis forces hx, yi = 0. 18. Every 7 × 7 matrix has at least one real eigenvalue. Solution note: True! The characteristic polynomial is degree 7. An odd degree polynomial always has at least one root. T 19. Let V → V be a linear transformation, and suppose that ~x and ~y are linearly independent eigenvectors with different eigenvalues. Then ~x + ~y is NOT an eigenvector. Solution note: TRUE! Say T (~x) = k1 ~x and T (~y ) = k2 ~y . Suppose T (~x + ~y ) = k3 (~x + ~y ). Then T (~x + ~y ) = T (~x) + T (~y ) so k3 (~x + ~y ) = k1 ~x + k2 ~y . Rewriting, we have (k3 − k1 )~x + (k3 − k2 )~y = 0. Since ~x and ~y are linearly independent, this relation must be trivial so (k3 − k1 ) = (k3 − k2 ) = 0. This implies k1 = k2 = k3 . 20. If hx, yi = hx, zi for vectors x, y, z in an inner product space, then y − z is orthogonal to x. Solution note: True: 0 = hx, yi−hx, zi = hx, y −zi, so the x and y −z are orthogonal. 21. For any matrix A, the system AT A = AT ~b is consistent. Solution note: True! The solutions are the least squares solutions.   2 0  22. If A is the B = (~v1 , ~v2 , ~v3 , ~v4 ) matrix of a transformation T and  1 are the B-coordinates 0 of ~x, then T (~x) = 2~v1 + ~v3 . Solution note: False!! By definition   of B-matrix, we know [T ]B [~x]B = [T (~x)]B = 2   0  T (~v1 ) T (~v2 ) T (~v3 ) T (~v4 )  1 = 2T (~v1 ) + T (~v3 ). This does not have to equal 0 2~v1 + ~v3 . The zero transformation for T is an explicit counterexample. 23. If A is the B = (~v1 , ~v2 , ~v3 , ~v4 ) matrix of a transformation T and T (~v3 ) = ~v1 + ~v3 , then A~e3 = ~e1 + ~e3 . Solution note: True. The third column of A tells us the B-coordinates of T (~v3 ). This should be [1 0 1 0]T . Also the third column of A is A~e3 . 24. If an 5 × 5 matrix P has eigenvalues 1, 2, 4, 8 and 16, then P is similar to a diagonal matrix. Solution note: Yes! There are 5 different eigenvalues and the matrix is size 5 × 5. So the geometric multiplicity of each is (at least) 1, and the sum is (at least) 5. So the matrix is diagonalizable by the theorem. 25. The functions sin x and cos x are orthogonal in the inner product defined by hf, gi = 1 2π Rπ −π f gdx. Solution note: True. Check hsin x, cos xi = 0. This is easy since sin x cos x is an odd function. 26. Suppose we have an inner product space V and w ~ and ~v are orthonormal vectors in V . Then for any f ∈ V , the element hw, f iw + hv, f iv is the closest vector to f in the span of v and w. Solution note: True! This is the formula for the projection of f onto the span of the {~v , w} ~ (because they are an orthonormal basis!). The projection is the closest vector. 27. In any inner product space, ||f || = hf, f i for all f . Solution note: False! Must square root! 2×2 as an inner product space with the inner product hA, Bi = trace AT B. Then 28. Consider  R  p a b || = (a2 + b2 + c2 + d2 ). || c d Solution note: True. Compute ||A|| = s  2  a + b2 − trace . − c2 + d2 p p hA, Ai = trace(AT A) =     0 1 1 0 29. The matrices and are orthonormal in the inner product hA, Bi = trace AT B 1 0 0 1 on R2×2 . Solution note: False. They are perpendicular (orthogonal) but not of length one. √ Each has ||A|| = 2. 30. If f and g are elements in an inner product space satisfying ||f || = 2, ||g|| = 4 and ||f +g|| = 5, then it is possible to find the exact value of hf, gi Solution note: True: 52 = ||f + g||2 = hf + g, f + gi = ||f ||2 + 2hf, gi + ||g||2 = 22 + 2(hf, gi) + 42 . This can be solved for hf, gi, which is 5/2. 31. If (~ for the subspace V of Rn and ~b ∈ V , thenthe least squares solutions v1 , . . . , vd ) is a basis  of ~v1 ~v2 . . . ~vd ~x = ~b are exact solutions to ~v1 ~v2 . . . ~vd ~x = ~b.  Solution note: True: since ~b ∈ V , it is in the span of the columns of ~v1 ~v2 . . . So it is consistent, so the least squares solutions are actual solutions.  ~vd . n ⊥ ~ 32. If (~v1 , . . . , vd) is a basis for the  subspace V of R and b ∈ V , then the only least squares ~ solutions of ~v1 ~v2 . . . ~vd ~x = b is the zero vector. Solution note: True. Because ~b ∈ V ⊥ , the projection of ~b to V is zero. This means that the least squares solutions are the actual solutions to A~x = ~0. Since the columns of A are a basis, they are linearly independent, which means that A is invertible. So the only solution of A~x = ~0 is 0. 33. Suppose a is an eigenvalue of an invertible matrix A. Then a−1 is an eigenvalue of A−1 . Solution note: True! We have A~v = a~v . Apply A−1 to both sides and divide by a (note it is not zero since an invertible matrix has non-zero eigenvalues). This gives A−1~v = a1 ~v . 34. If A is upper triangular, then A is diagonalizable. Solution note: False! The eigenvalues are the elements of the diagonal. But we don’t know  the geometric multiplicities sum up to the desired value. A counterexample  that 0 1 is 0 0 35. Every lower triangular matrix with distinct eigenvalues has an eigenbasis. Solution note: True! Since the matrix is lower triangular, all eigenvalues are real. Since they are distinct, the geometric multiplicity of each is one, and these sum to the size of the matrix. 36. There are no surjective maps P4 → R10 . Solution note: True! Rank nullity! Dimension of P4 is 5, so the largest the dimension of the image can be is 5, never 10. 37. There are no injective maps P14 → R10 . Solution note: True! Rank nullity. Kernel has dimenion at least 5. 2×2 defined by hA, Bi = trace(AT B). Then the matrices 38.  Consider  the inner  product on R 0 0 1 0 and are orthonormal. 0 0 0 1       0 0 0 0 0 0 Solution note: True! Compute h , i = trace( ) = 1, 0 1 0 1 0 1       1 0 1 0 1 0 h , i = trace( ) = 1, 0 0 0 0 0 0       0 0 1 0 0 0 and h , i = trace( ) = 0. So these are orthonormal. 0 1 0 0 0 0   a b 39. Using the inner product from the previous problem, the closest diagonal matrix to is c d   a 0 . 0 d Solution note: True! We need to compute the projection onto the space of diagonal matrices. Since we already found an orthonormal basis in 38, this is easy. Call this basis (E11 , E22 ). The projection is hA, E11 iE11 + hA, E22 iE22 . Computing this coeeficients we have hA, E11 i = trac(AT E11 ) = a and hA, E22 i = trac(AT E22 ) = d. So the closest matrix is aE11 + dE22 . 40. There is a matrix which has determinant 6 and trace 5. Solution note: True: A diagonal 2 × 2 with 2 and 3 on the diagonal. 41. If a 2 × 2 matrix A has characteristic polynomial x2 + bx + c, then it has an eigenvalue of algebraic multiplicity two if and only if b2 = 4c. Solution note: True! Having an eigenvalue of multiplicity two means that the characteristic polynomial has one double root. According to the quadratic formula, this happens if and only if b2 − 4c = 0. The eigenvalue in this case is the root −b/2. 42. A 2 × 2 matrix A has no real eigenvalues if and only if (trace A)2 < 4 det A. Solution note: True! The char poly is x2 + bX + c where −b is the trace and c is the determinant. Again, using the quadratic formula, we see that this has no real roots if and only if b2 − 4c < 0. 43. If some eigenspace of an n × n matrix A has dimension n, then A is a scalar multiple of the identity matrix. Solution note: True! If A has an eigenspace of dimension n, then the eigenspace is all of Rn . This means that A~v = k~x for all vectors ~x ∈ Rn . So the map is just scalar multiplication by k and the matrix A must be kIn . 44. Let S be an orthogonal 3 × 3 matrix. The linear transformation R3×3 7→ R3×3 sending X 7→ S T XS is invertible. Solution note: True! The inverse map is Y 7→ SY S T . Note that S T (SY S T )S = Y and S(S T XS)S T = X for all X and all Y . 45. Let S be an invertible 3 × 3 matrix. The only eigenvalues of the linear transformation R3×3 7→ R3×3 sending X → 7 S −1 XS are 0 and 1. Solution note: False! This map is invertible, so zero is not an eigenvalue. 46. For any n × n matrix A, the determinant of kA is k n det A. Solution note: True. Multiplinearity of determinant. 47. There exists an orthogonal matrix with eigenvalues 3, 2 and 1. Solution note: False! Orthogonal matrices can have eigenvalues ±1 only! Or no eigenvalues, like a rotation. 48. There exists a symmetric matrix with no real eigenvalues. 49. Let S be an orthogonal matrix and D be diagonal of the same size as S. Then S −1 DS is symmetric. Solution note: TRUE! We check that S −1 DS = (S −1 DS)T . Note that S −1 = S T . So (S −1 DS)T = S T DT (S T )T = S −1 DS. 50. If a square matrix B has an orthonormal eigenbasis, then B is symmetric. Solution note: True! This is (one direction of) the spectral theorem. Or, you can interpret as the same as the previous problem. 51. If an 7×7 matrix Q has eigenvalues 1 of geometric multiplicity 3 and 2 of geometric multiplicity 4, then Q is invertible. Solution note: True! The geometric multiplicities sum to 3 + 4 = 7, so the matrix has an eigenbasis, which means that it is diagonal. 52. There is a 10 by 10 matrix with eigenvalues 1, 2, . . . , 10. Solution note: True! Just take the diagonal matrix with 1 − 10 on the diagonal. 53. There is noninvertible 10 by 10 matrix with eigenvalues 1, 2, . . . , 10. Solution note: False! Non-invertible would mean that 0 is an eigenvalue. But there can be at most 10 eigenvalues for a 10 by 10 matrix, and we know that they are 1–10 in this case. None is zero. 54. There is non-diagonalizable 10 by 10 matrix with eigenvalues 1, 2, . . . , 5, each of algebraic multiplicity 2.  1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0  0 0 2 1 0 0 0 0 0  0 0 0 2 0 0 0 0 0  0 0 0 0 3 1 0 0 0 Solution note: True! Take  0 0 0 0 0 3 0 0 0  0 0 0 0 0 0 4 1 0  0 0 0 0 0 0 0 4 0  0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 rank-nullity that all the geometric multiplicities are 1.  0 0  0  0  0  . We can compute using 0  0  0  1 5 55. There is 10 by 10 matrix with eigenvalues 1, 2, . . . , 5, each of geometric multiplicity 2, which does not have an eigenbasis. Solution note: False! The sum of the geometric multiplicities in this case is 10, so we must have an eigenbasis. 56. There is non-zero 10 by 10 matrix with  0 0 0 0 0 0 0 0  0 0 0 1  0 0 0 0  0 0 0 0 Solution note: Here is it:  0 0 0 0  0 0 0 0  0 0 0 0  0 0 0 0 0 0 0 0 an eigenvalue 0 of algebraic multiplicity 10.  0 0 0 0 0 0 0 0 0 0 0 0  0 0 0 0 0 0  0 0 0 0 0 0  0 1 0 0 0 0 . 0 0 0 0 0 0  0 0 0 1 0 0  0 0 0 0 0 0  0 0 0 0 0 1 0 0 0 0 0 0 57. There is non-zero 10 by 10 matrix with an eigenvalue 0 of geometric multiplicity 10. Solution note: False. If the geometric multiplicity is 10, then the matrix is diagonalizable. But since the only eigenvalue is 0, it would be diaogonalizable to the zero matrix. This is impossible-the only matrix similar to the zero matrix is the zero matrix. 58. There is a 10 by 10 matrix with an eigenvalue λ of geometric multiplicity 5 and algebraic multiplicity 2. Solution note: False! Gemu ¡ almu for all eigenvalues! 59. The only matrix similar to the zero matrix is the zero matrix itself. Solution note: True! S −1 0S = 0. 60. The only matrix similar to the identity matrix is the identity matrix itself. Solution note: True! S −1 In S = In . 61. There is a non-diagonal matrix similar to kIn . Solution note: False! S −1 kIn S = kIn . 62. A matrix has an eigenbasis if and only if it is diagonalizable. Solution note: Yes, this is a basic theorem. 63. A matrix has an eigenbasis if and only if all eigenvalues has geometric multiplicity one. Solution note: False! This is silly. The identity matrix A2 has an eigenbasis (any basis for R2 is an eigenbasis) but the geometric mutltiplicity 2 of the eigenvalue 1 is two,