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Math 217: True False Practice Professor Karen Smith 1. a Square Matrix Is Invertible If and Only If Zero Is Not an Eigenvalue. Solution Note: True

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(c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 217: True False Practice Professor Karen Smith 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: False. This is silly. Let A = B = 5I2. Then the eigenvalues of AB are 25. 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. 5. If B is the B-matrix of some linear transformation V !T V . Then for all ~v 2 V , we have B[~v]B = [T (~v)]B. Solution note: True. This is the definition of B-matrix. 21 2 33 T 6. Suppose 40 2 05 is the matrix of a transformation V ! V with respect to some basis 0 0 1 B = (f1; f2; f3). 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. 21 2 33 T 7. Suppose 40 2 05 is the matrix of a transformation V ! V with respect to some basis 0 0 1 B = (f1; f2; f3). Then T (f1 + f2 + f3) is 6f1 + 2f2 + f3. 213 Solution note: TRUE! The B-coordinates of f1 + f2 + f3 are 415. To get the B- 1 21 2 33 coordinates of T (f1 + f2 + f3) we just multiply by the matrix [T ]B = 40 2 05 to 0 0 1 263 get 425. 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. 2 2 9. The only rotation R ! R 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. T 2×2 11. The map assigning hA; Bi to trace(AB ) is an inner product on the space of all R 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! T T 2×2 13. Both hA; Bii = trace A B and hA; Bi = trace AB define an inner product on R . 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 p 16. Let u and v be any two orthonormal vectors in an inner product space. Then jju − vjj = 2. Solution note: True. jju−vjj 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. 19. Let V !T 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. 223 607 22. If A is the B = (~v1;~v2;~v3;~v4) matrix of a transformation T and 6 7 are the B-coordinates 415 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 = 223 607 T (~v1) T (~v2) T (~v3) T (~v4) 6 7 = 2T (~v1) + T (~v3): This does not have to equal 415 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 T should be [1 0 1 0] . 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. 1 R π 25. The functions sin x and cos x are orthogonal in the inner product defined by hf; gi = 2π −π fgdx. 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 2 V , the element hw; fiw + hv; fiv 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 f~v; ~wg (because they are an orthonormal basis!). The projection is the closest vector. 27. In any inner product space, jjfjj = hf; fi for all f. Solution note: False! Must square root! 2×2 T 28. Consider R as an inner product space with the inner product hA; Bi = trace A B. Then a b jj jj = p(a2 + b2 + c2 + d2). c d Solution note: True. Compute jjAjj = phA; Ai = ptrace(AT A) = s a2 + b2 − trace . − c2 + d2 0 1 1 0 29. The matrices and are orthonormal in the inner product hA; Bi = trace AT B 1 0 0 1 2×2 on R . Solution note: False.p They are perpendicular (orthogonal) but not of length one. Each has jjAjj = 2. 30. If f and g are elements in an inner product space satisfying jjfjj = 2; jjgjj = 4 and jjf +gjj = 5, then it is possible to find the exact value of hf; gi Solution note: True: 52 = jjf + gjj2 = hf + g; f + gi = jjfjj2 + 2hf; gi + jjgjj2 = 22 + 2(hf; gi) + 42.
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  • Sam Roweis' Notes on Matrix Identities

    Sam Roweis' Notes on Matrix Identities

    matrix identities sam roweis (revised June 1999) note that a,b,c and A,B,C do not depend on X,Y,x,y or z 0.1 basic formulae A(B + C) = AB + AC (1a) (A + B)T = AT + BT (1b) (AB)T = BT AT (1c) 1 1 1 if individual inverses exist (AB)− = B− A− (1d) 1 T T 1 (A− ) = (A )− (1e) 0.2 trace, determinant and rank AB = A B (2a) j j j jj j 1 1 A− = (2b) j j A j j A = evals (2c) j j Y Tr [A] = evals (2d) X if the cyclic products are well defined; Tr [ABC :::] = Tr [BC ::: A] = Tr [C ::: AB] = ::: (2e) rank [A] = rank AT A = rank AAT (2f) biggest eval condition number = γ = r (2g) smallest eval derivatives of scalar forms with respect to scalars, vectors, or matricies are indexed in the obvious way. similarly, the indexing for derivatives of vectors and matrices with respect to scalars is straightforward. 1 0.3 derivatives of traces @Tr [X] = I (3a) @X @Tr [XA] @Tr [AX] = = AT (3b) @X @X @Tr XT A @Tr AXT = = A (3c) @X @X @Tr XT AX = (A + AT )X (3d) @X 1 @Tr X− A 1 T 1 = X− A X− (3e) @X − 0.4 derivatives of determinants @ AXB 1 T T 1 j j = AXB (X− ) = AXB (X )− (4a) @X j j j j @ ln X j j = (X 1)T = (XT ) 1 (4b) @X − − @ ln X(z) 1 @X j j = Tr X− (4c) @z @z T @ X AX T T 1 T T T 1 j j = X AX (AX(X AX)− + A X(X A X)− ) (4d) @X j j 0.5 derivatives of scalar forms @(aT x) @(xT a) = = a (5a) @x @x @(xT Ax) = (A + AT )x (5b) @x @(aT Xb) = abT (5c) @X @(aT XT b) = baT (5d) @X @(aT Xa) @(aT XT a) = = aaT (5e) @X @X @(aT XT CXb) = CT XabT + CXbaT (5f) @X @ (Xa + b)T C(Xa + b) = (C + CT )(Xa + b)aT (5g) @X 2 the derivative of one vector y with respect to another vector x is a matrix whose (i; j)th element is @y(j)=@x(i).