MATH 431: FIRST MIDTERM. Thursday, October 3, 2013.

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1 MATH 431: FIRST MIDTERM Thursday, October 3, 213. (1) An inner product on the space of matrices. Let V be the vector space of 2 2 real matrices (that is, the algebra Mat 2 (R), but without the mulitiplicative structure). Recall that if W V is a linear subspace of a vector space V, and V has an inner product (denoted v w), then the orthogonal complement W of W inside V is defined by: W := {v V v w =, w W} and every vector v V decomposes uniquely as v = w + u where w W and u W. (These are the orthogonal projections of v to W and W, respectively.) (a) Show that the mapping V V R (A, B) A B := Tr(AB ) is bilinear, symmetric and positive definite, where B denotes the transpose of B. We shall call this the inner product on the vector space of matrices. Solution: Let s, t R and A, B, C Mat 2 (R). Our lives will be easier if we observe the recall the following from linear algebra, which can be proved easily enough in the 2 2 cases by expanding the formulas: (i) tr(ab) = tr(ba). (ii) tr(a ) = tr(a). (For a square matrix A and A have the same diagonal entries). Date: October 9,

2 2 MATH 431: FIRST MIDTERM Symmetry. (A, B) = tr(ab ) = tr ( (AB ) ) By (ii) = tr ( (B ) A ) = tr(ba ) = (B, A) Bilinear First, note that the trace is a linear map. If A and B are two matrices, then since matrix addition is entry-wise, A + B has diagonal elements {a ii + b ii }. Similarly, sa + B has entries {sa ii + b ii }. Thus in general tr(sa + B) = i sa ii + b ii = s i a ii + j b jj = s tr(a) + tr(b) It is linear in the first factor since (sa + B, C) = tr(sac + BC ) = s tr(ac ) + tr(bc ) = s(a, B) + (A, C) It is then automatically linear in the second factor by symmetry: (A, sb + C) = (sb + C, A) = s(b, A) + (C, A) = s(a, B) + (A, C) Positive Definite. Many people misunderstand waht it means to be positive definite. The standard dot product on R 2 is positive definite, but it is not true that X Y > for all X, Y nonzero in R 2. For example, X = e 1, Y = e 1. Then e 1 e 1 = 1. Instead, positive definiteness requires that if A is nonzero, the product with itself is positive. That is we require (A, A) >. But (AA ) ii = n j=1 a ij(a ji ). Since (a ji ) = a ij by the definition of transpose, (AA ) ij = n j=1 a2 ij. In particular, since at least one entry of A is nonzero if A, we have tr(aa ) > by the positivity of squares of real numbers. (b) Find an orthonormal basis of the inner product space (V, ).

3 MATH 431: FIRST MIDTERM 3 Solution: The simplest thing to do is to check the basis corresponding {( to the ) standard orthonormal ( basis )} of R This would be,,,. Direct computation shows that this set is orthonormal with 1 1 respect to ourbilinear form. It is clear that this set is linearly independent and spans (we showed this in HW 1). (c) Show that the mapping Mat 2 (R) [ ] x y x + iy y x C E embeds the algebra of complex numbers into the algebra of 2 2 real matrices. That is, show that this mapping is one-to-one, linear and takes multiplication of complex numbers into matrix multiplication. Let { [ ] x, } x y W := E (C) = y R y x be the space of matrices corresponding to complex numbers. Solution: We showed in the previous homework that for z C, E (z) corresponds to multiplication by a complex number. You had to show this on the test, but if you want the solution, please see the solutions for that homework. We have to check that this map is one-to-one and linear. The map is one-to-one since we can read the pre-image z from a matrix in W off of the first column. Thus if E (z) = for z, u C, we have that the first columns are equal, and so Re(z) = Re(u) and Im(z) = Im(u), which implies z = u. Now let us show that the map is linear. Let λ, x, y, a, b R. λx + a λy b E (λ(x + iy) + (a + ib)) = λy + b λx + a x y a b = λ + y x b a = λe (x + iy) + E (a + ib) (d) Compute the orthogonal complement W of W V.

4 4 MATH 431: FIRST MIDTERM a b Solution: Let A = be an element of W c d. A x y general element of W is given by X =. By y x definition of W, tr(ax ) =. ([ ] [ ]) a b x y tr(ax ) = tr c d y x ([ ]) ax by ay + bx = tr cx dy cy + dx = ax by + cy + dx = Since ax by + cy + dx = for all x, y R, we can substitute x = 1, y = to find a = d, and [ x = ], y = 1 a b to find b = c. Thus A has the form A =. Since b a ax by +by +ax = for all a, b, x, y R, we see that a = d and b = c are the only conditions required for A to be in W. (e) If A V, then compute the orthogonal decomposition A = A + + A where A + W, A W. Solution: This can be solved using a linear system of equations as in the previous homework. In this case, we get a b = 1 a + d b c + 1 a d b + c c d 2 c b a + d 2 c + b d a (2) A periodic collineation. Consider the collineation φ of P 2 defined by the matrix in homogeneous coordinates. Prove or disprove: (a) For every q P 2, φ(q) q. (Remember: points in P 2 are equivalence classes of vectors, and not themselves vectors.)

5 MATH 431: FIRST MIDTERM 5 Solution: False. Let A be the linear map with matrix above, and let φ be the corresponding collineation. Then a fixed point of φ is just a nonzero eigenvector of A, since if Av = λv, we have v λv in P 2. In particular Ae 3 = e 3, so [e 3 ] (the equivalence class of e 3 in P 2 ) is a fixed point of φ. (b) The (projective) line l := {[X : Y : ] P 2 (X, Y ) (, )} P 2 is invariant under φ. Solution: True. l is the line at infinity and is the projectivization of the plane defined by the equation z =. We calculate X Y = Y X Thus φ takes a point on l to another point on l and so l is invariant under φ. (c) If q l, then φ(q) q but φ 2 (q) = q. (Here φ 2 means the composition φ φ.) Solution: True. For a point v = X Y on the plane z =, Av = Y X, and A 2 (v) = X Y = v. Since v v, φ 2 ([v]) = [v]. (d) Find the smallest n > such that the n-fold composition φ n equals the identity map on P 2, where φ n := n times {}}{ φ φ φ. Solution: A collineation of P 2 is the identity if and only if the corresponding matrix is a scalar multiple of the identity. You can see this rather easily. If A is the matrix corresponding to your collineation, calculate the image under A of any vector that is not an eigenvector. Then the collineation does not fix the point corresponding to that vector. With A above, it is easy to calculate that A, A 2

6 6 MATH 431: FIRST MIDTERM and A 3 are not multiples of the identity, but A 4 is the identity matrix. Thus n = 4.

Review problems for MA 54, Fall 2004.

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