MATH 33A LECTURE 3 PRACTICE MIDTERM I

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1 MATH A LECTURE PRACTICE MIDTERM I Please note: Show your work Correct answers not accompanied by sufficent explanations will receive little or no credit (except on multiple-choice problems) Please call one of the proctors if you have any questions about a problem No calculators computers PDAs cell phones or other devices will be permitted If you have a question about the grading or believe that a problem has been graded incorrectly you must bring it to the attention of your professor within weeks of the exam # # # #4 #5 Total TA name (circle): Your section meets (circle): Ioannis Lagkas-Nikolos Tuesday Thursday Jaehoon Lee Austin Christian Name: Student ID: Signature: By signing above I certify that I am the person whose name and student ID appears on this page Copyright c 6 UC Regents/Dimitri Shlyakhtenko No part of this exam may be reproduced by any means including electronically

2 MATH A LECTURE PRACTICE MIDTERM I Problem (True/False pt each) Mark your answers by filling in the appropriate box next to each question (no explanations are necessary on this problem) (a T F ) There exists a matrix A so that A = A A = but A A A = NO Since A is ker A is a subspace of R and can have dimension or If dim ker A = A = ; but we are given that A = So ker A can be dimension or dim If ker A had dimension the nullity of A would be zero so the rank of A would be by the rank-nullity theorem So A would invertible but then A would also be invertible and so could not be zero So ker A must be one-dimensional By rank-nullity also im A is then also one-dimensional Let s say that {v} is a basis for im A Since Av is in the image of A Av must be a multiple of v: Av = αv If α = we would get that A v = A A Av = A Aαv = Aα v = α v = contradicting A = So αmust be zero If α = then Av = Let w be any vector Then Aw is in the image of A thus a multiple of v Then A Aw must be a mutliple of Av which is zero So A = again a contradiction Thus such an A cannot exist (b T F ) There are two matrices A B so that AB = BA YES Take for example A = B = Then Ae = Ae = e Be = e so ABe = Ae = e but BAe = because Ae = So ABe = BAe so AB = BA (c T F ) If A is an n n matrix with rank A = then the system of equations Ax = always has exactly one solution NO The solution would be unique exactly when rank A = n; so if n > there would be more than one solution (d T F ) If A is an n m matrix and im A = R n then n m YES The image is spanned by m vectors Ae Ae m so by assumption R n is spanned by these m vectors Since dim R n = n it must be that m n

3 MATH A LECTURE PRACTICE MIDTERM I (e T F ) Let A B C be matricse If AB = I and AC = I then C = B YES If AB = I it follows that im A = R so A has rank and so is invertible Multiplying AB = I on both sides on the left by A gives A AB = A so B = A Multiplying AC = I on both sides by A on the left gives similarly that C = A Thus C = B (f T F ) The intersection of two subspaces may not be a subspace FALSE (see book) (g T F ) If ker A = im A for some square n n matrix A then n must be even TRUE By rank-nullity n = dim ker A+dim ima so if im A = ker A then n = dim ker A+ dim ker A = dim ker A so must be even (h T F ) There is no matrix A so that A = A FALSE The matrix of any reflection satisfies this property (i T F ) The kernel of a matrix is a subspace TRUE (see book) (j T F ) Every subspace is the image of some matrix TRUE Suppose that the subspace is the span of v v v m Then it is the image of the matrix with columns v v m

4 MATH A LECTURE PRACTICE MIDTERM I 4 Problem ( pts) Let A = Find the rank of A a basis for the image and a basis for the kernel of A Solution The reduced row echelon form is 9 It follows that columns 4 6 of the original matrix form a basis for the image Thus a possible basis is A vector a a a 7 is in the kernel if a = t 9 s a = t+ s a = t a 4 = s a 5 = r a 6 = s so that a basis for the kernel is 9

5 MATH A LECTURE PRACTICE MIDTERM I 5 Problem ( pts) Let T be o the rotation about the y axis clockwise in the xz plane as seen from the positive y axis Find the matrix of T Solution Restricting to the xz plane we see that T e = cos o e + sin o e = e + e T e = e + e Also T e = e Thus the matrix is

6 MATH A LECTURE PRACTICE MIDTERM I 6 Problem 4 ( pts) Find all matrices A satisfying A = A Solution Since det(a ) = (det A) we know that det A = det(a ) = (det A) so that det A det A = So det A = ± Suppose that det A = Then [ a b A = c d ] A = det A [ ] [ d b d b = c a c a Since A = A we get that a = d b = b c = c Thus b = c = and a = d Since det A = we get that ad = so a = Thus the only possibilities are a = d = ± b = c = : [ ] [ ] Clearly both of these matrices satisfy A = A Suppose now that det A = Then [ ] a b A = A c d = [ ] [ d b d b = det A c a c a Thus a = d d = a Since det A = we get that ad bc = so either c = (which forces a = ) or b = (ad+)/c = ( a )/c Thus A could be [ ] [ ] or [ a ( a A = ] )/c c a in each case the verification that A = A being easy ] ]

7 MATH A LECTURE PRACTICE MIDTERM I 7 Problem 5 ( pts) Let V W be subspaces of R 4 given by: [ ] 4 V = ker and W = ker Find a basis for their intersection V W Solution The intersection is the set of all vectors v so that [ ] 4 v = and The resulting system of linear equations has the matrix A = whose reduced row echelon form is v = which means that the only possible vector v which is in both spaces is the zero vector Thus the basis for the intersection is the empty set