Chapters 5 & 6: Theory Review: Solutions Math 308 F Spring 2015

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1 Chapters 5 & 6: Theory Review: Solutions Math 308 F Spring 205. If A is a 3 3 triangular matrix, explain why det(a) is equal to the product of entries on the diagonal. If A is a lower triangular or diagonal matrix, a 0 0 A = a 2 a 22 0 a 3 a 32 a 33 we can compute det(a) using an expansion across the first row, so det(a) = a C + 0 C C 3 = a (a 22 a 33 0 a 32 ) = a a 22 a 33. If A is upper triangular, we could compute det(a) using an expansion across the first column, and get the same result OR we could notice that A T is a lower triangular matrix with the same entries as A on the diagonal, so by what we just said, det(a) = det(a T ) is the product of entries on the diagonal. Note that this formula holds for ANY n n triangular matrix, not just a 3 3 matrix. 2. If A is an n n matrix, what is the definition of (a) an eigenvalue of A? An eigenvalue of A is a scalar λ such that the equation Ax = λx has a nonzero solution. (b) an eigenvector of A? An eigenvector of A is a nonzero vector u such that Au = λu for some scalar λ (in other words, a nonzero solution to the above equation). (c) Explain how finding eigenvalues/vectors can be restated in terms of determinants and singular matrices. If we rewrite the equation Ax = λx as Ax λx = 0, and then factor out the x to get (A λi)x = 0, we can see that finding eigenvalues and eigenvectors is the same thing as finding scalars λ such that (A λi)x = 0 has nonzero solutions. But, this has nonzero solutions if and only if the matrix A λi is singular, which occurs if and only if det(a λi) = 0. So, to find eigenvalues, we can first find all the values of λ such that det(a λi) = 0, and then we can find eigenvectors by solving the equation (A λi)x = 0 (which is guaranteed to have a nonzero solution because A λi is singular).

2 3. If B is a basis for R n, explain what is meant by the notation x B. The vector x B is the vector of coordinates of x in the basis B. If B is the basis {u,..., u n }, then the notation means c x B =... c n x = c u + + c n u n. B 4. If B and B 2 are two bases for R n, what is the change of basis matrix from B to B 2? The change of basis matrix is an invertible n n matrix M that allows us to translate between the coordinates of x in one basis to the coordinates of x in another basis. Specifically, the change of basis matrix from B to B 2 is a matrix M such that x B2 = Mx B. There is a formula for the matrix M: if B = {u,..., u n } and B 2 = {v,..., v n }, let U be the matrix U = [u... u n ] and V be the matrix V = [v... v n ]. Then, the matrix M is given by M = V U. 5. Give an example of the following, or explain why such an example does not exist. (a) A nonsingular matrix A such that det(a) = det(a 2 ) Because det(a 2 ) = det(a A) = det(a) det(a) = det(a) 2, if det(a) = det(a) 2, we must have det(a) = or det(a) = 0. In order to be nonsingular, we must have det(a) 0. Therefore, any matrix whose determinant is one will satisfy this, so one example would be the identity matrix, [ ] 0 A = 0 and another would be the matrix A = [ ] 2.

3 (b) A nonsingular matrix A such that det(a) = 0 This is impossible. An nonsingular matrix must be invertible, meaning det(a) 0. (c) A nonsingular matrix A with only one eigenvalue One example of this is the identity matrix, [ ] 0 A =, 0 whose only eigenvalue is one. Any triangular matrix whose entries on the diagonal are equal and nonzero would also work, because the eigenvalues of a triangular matrix are just the numbers on the diagonal. (d) A singular matrix A with only one eigenvalue One example of this is the zero matrix, [ ] 0 0 A =, 0 0 whose only eigenvalue is 0. Any triangular matrix whose entries on the diagonal are all 0 would also work, because the eigenvalues of a triangular matrix are just the numbers on the diagonal. (e) An n n matrix A that has less than n eigenvalues See parts (c) and (d). Any matrix with at least one repeated eigenvalue will have less than n eigenvalues. (f) An n n matrix A with more than n eigenvalues This is impossible. The eigenvalues are the roots of the characteristic polynomial det(a λi). If A is an n n matrix, this is a degree n polynomial, so it can have at most n distinct roots. 6. (This problem is part of a problem in an old exam.) Consider a 4 4 matrix A such that the following eigenvectors of A form a basis of R 4 : ,,, eigenvalues: 2

4 (a) Find a basis for null(a 2I). Explain why your answer is correct. The eigenspace of 2 is precisely the nullspace of A 2I, and the eigenvalue 2 has multiplicity, meaning the dimension of the eigenspace must actually be. Therefore, a basis for null(a 2I) is the eigenvector corresponding to 2, 0. (b) What is rank(a I)? Justify your answer. The eigenspace of has dimension 2 because the vectors corresponding to the eigenvalue are linearly independent. Therefore, nullity(a I) = 2, so by the rank-nullity theorem, rank(a I) = 2. (c) If x = 3, find A7 x. 3 We can write x as a linear combination of eigenvectors: 0 x = so A 7 x = A A = ( )7 = 0 2 = For each statement below, determine if it is True or False. Justify your answer. The justification is the most important part! (a) If A and B are n n matrices, then AB is singular if and only if A or B is singular. TRUE. AB is singular if and only if det(ab) = 0. But, det(ab) = det(a) det(b), so this is zero if any only if det(a) = 0 or det(b) = 0, which holds if and only if A or B is singular.

5 (b) If T : R n R n is a linear transformation with ker(t ) {0}, then the matrix A such that T (x) = Ax has det(a) = 0. TRUE. This is part of the Big Theorem. If ker(t ) {0}, then null(a) {0}, so there is some nonzero solution to the equation Ax = 0. But, that means the columns of A are linearly dependent, so A is singular, so det(a) = 0. (c) If A and B are row equivalent matrices, then det(a) = det(b). FALSE. If A = [ ] 0, B = 0 [ ] then A and B are row equivalent, but det(a) = det(b) = 2. (d) If det(a) = det(b) and A is a nonsingular matrix, then A and B are row equivalent matrices. TRUE. If A is nonsingular, det(a) 0 and A row reduces to the identity matrix. But, because 0 det(a) = det(b), we know B is also nonsingular, so B also row reduces to the identity matrix. Therefore, A and B are row equivalent (they reduce to the same matrix). (e) If det(a) = 0, then one of the columns of A can be written as a linear combination of the others. TRUE. This is part of the Big Theorem. If det(a) = 0, then the columns of A are not linearly independent, so one must be a linear combination of the other columns. (f) If det(a) = 0, then 0 is an eigenvalue of A. TRUE. If 0 = det(a) = det(a 0 I), then 0 is an eigenvalue of A because is satisfies the equation det(a λi) = 0. We could also justify this with the Big Theorem: if det(a) = 0, then A is singular, so the Big Theorem tells us 0 must be an eigenvalue of A. (g) If 0 is an eigenvalue of A, then the linear transformation T (x) = Ax is not onto. TRUE. If 0 is an eigenvalue of A, by the Big Theorem, the columns of A do not span R n. Therefore, the linear transformation T is not onto. (h) If λ is an eigenvalue of A, the eigenspace of λ is equal to null(a λi). TRUE. The eigenspace of λ is defined to be all vectors x that satisfy the equation Ax = λx (eigenspace = all eigenvectors plus the zero vector). But,

6 we can rewrite this equation as (A λi)x = 0 (see problem 2(c)), and the set of all solutions to this equation is exactly null(a λi). (i) If λ is an eigenvalue of A, then any vector u such that Au = λu is an eigenvector. FALSE. The zero vector satisfies A0 = λ0, but eigenvectors are required to be nonzero. (j) If B and B 2 are two bases for R n, then the change of basis matrix from B to B 2 is invertible. TRUE. The change of basis matrix is ALWAYS invertible. We could see this from the formula: if B = {u,..., u n } and B 2 = {v,..., v n }, let U be the matrix U = [u... u n ] and V be the matrix V = [v... v n ]. Then, the matrix change of basis matrix is given by V U, which has an inverse U V, so is invertible. Note that each of U and V are invertible because their columns form a basis for R n, so are linearly independent and spanning. (k) If M is the change of basis matrix from B to B 2 and N is the change of basis matrix from B 2 and B 3, then MN is the change of basis matrix from B and B 3. FALSE. The problem says that x B2 = Mx B and x B3 = Nx B2. Substituting the first equation in to the second, we see that x B3 = N(Mx B ) = NMx B so the change of basis matrix from B and B 3 is NM.

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