Math 115A HW4 Solutions University of California, Los Angeles. 5 2i 6 + 4i. (5 2i)7i (6 + 4i)( 3 + i) = 35i + 14 ( 22 6i) = i.
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1 Math 5A HW4 Solutions September 5, 202 University of California, Los Angeles Problem 4..3b Calculate the determinant, 5 2i 6 + 4i 3 + i 7i Solution: The textbook s instructions give us, (5 2i)7i (6 + 4i)( 3 + i) = 35i + 4 ( 22 6i) = i 2 Problem 4..7 Prove that det(a t ) = det(a) for any A M 2 2 (F ). Proof: Write out, a b A = c d so that, A t a c = b d Then we have det(a) = ad bc = ad cb = det(a t ). 3 Problem 4..9 Prove that det(ab) = det(a) det(b) for any A, B M 2 2 (F ). Proof: Since we are just dealing with 2 2 matrices, we can just write out what AB is and take its determinant, showing that it will be the same as det(a) det(b). We have, AB = The determinant of this is: a a 2 b b 2 a b = + a 2 b 2 a b 2 + a 2 b 22 a 2 a 22 b 2 b 22 a 2 b + a 22 b 2 a 2 b 2 + a 22 b 22 (a b + a 2 b 2 ) (a 2 b 2 + a 22 b 22 ) (a b 2 + a 2 b 22 ) (a 2 b + a 22 b 2 ) ()
2 But we also have, Multiplying this out clearly gives (). det(a) det(b) = (a a 22 a 2 a 2 ) (b b 22 b 2 b 2 ) 4 Problem Evaluate the determinant along the third row: 0 + i 2 2i 0 i 3 4i 0 Solution: We proceed carefully, taking the correct signs into account, yielding, + i i 3 det 4i det + 0 det = = 22 0 i 2i i 2i 0 5 Problem Calculate the determinant, Solution: There are a bunch of zeros in the third row and third column. The last problem asked us to calculate the determinant along the third row, so let s do the third column. Again, we check to see when to multiply by and we have (since there are 0 s in the second and third entries of the last column), 6 Problem det 5 6 = Prove that an upper triangular n n matrix is invertible if and only if all its diagonal entries are nonzero. Proof: Let Ω denote the set of all invertible matrices. Suppose A M n n (F ) and is triangular. We must first prove the following, 2
3 Lemma: If A is triangular, then, det(a) = (That is, the determinant is the product of the diagonal entries.) n A ii (2) Proof: If A is, the determinant is simply the A entry. Suppose that (2) holds for all n n matrices. Suppose then that A is n n and suppose also, without loss of generality, that A is upper triangular (if not, then by theorem 4.8, det(a t ) = det(a)). We have a matrix that looks something like this, A... i= 0 A nn Notice that I used a to denote the nonzero entries above the diagonal. This is because these entries will become largely irrelevant momentarily. Well, let s expand along the first column (since the determinant does not depend on how we calculate it). Then what results is, det(a) = A det A where A denotes the matrix consisting of all entries A ij where 2 i, j n. But the A matrix is n n, so its determinant follows (2), except the limit is n instead of n. Thus, what results is, n det(a) = A which is exactly (2). The result follows by induction. Now, by the lemma, if A Ω, it suffices to show that det(a) 0. But this follows by a corollary to theorem 4.7 in the textbook. Thus, no A ii entry can possibly be zero. On the other hand, if all A ii 0, then by the lemma, det(a) 0. By the same corollary, A Ω. i= A ii 7 Problem 4.3. Prove that if M is skew-symmetric and n is odd, then M is not invertible. What happens if n is even? Proof: Well, M is skew-symmetric, so, M t = M. But we also know that det(m) = det(m t ), so det(m) = det( M). But also, det( M) = ( ) n det M. Since n is odd, we conclude that det M = det M, which is not possible unless det M = 0. Ergo, M / Ω (notation from previous problem). 3
4 Notice, however, that if n is even, M may be invertible. Consider, for instance, It s determinant is, so it is invertible Problem 4.3.3b Prove that if Q is a unitary matrix, then det(q) =. Proof: By the previous result (which you were not required to show), det(m) = det(m). But we also know that, Therefore, we have, det(qq ) = det(i) But this is, det(q) det(q ) = det(q)det(q t ) = det(q)det(q) = det(q) = Done. If the last step isn t clear, if c = a + bi, then c c = (a + bi)(a bi) = a 2 + b 2 = c. 9 Problem 5..3c See text for instructions. The matrix given is, i 2 i Solution: Let A be the matrix. Then det(a λi) = λ 2. Hence, the eigenvalues are and. Now, we find a vector b such that, It is apparent that a vector that will work is, i b = 0 2 i i + 4
5 We do the same thing for the eigenvalue and get the following eigenvector, What emerges then is that a basis for C 2 is this, i { B = i + [, i since these vectors are clearly linearly independent. The matrix Q that we want then is, And we are done. [ ] i + ]} 0 Problem 5..4 For any square matrix A, prove that A and A t have the same characteristic polynomial (and hence the same eigenvalues). Proof: We have often used the property that det A = det A t and we will use this again. characteristic polynomial, P (λ) for a matrix A is given by, But notice, P (λ) = det(a λi) = The det(a λi) = det((a λi) t ) Since I is diagonal, (A λi) t = A t λi, so we get the desired result, det(a λi) = det(a t λi) 5
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