MATH 294 - Reference Sheet

MATH 294 - Reference Sheet - Chapter 1 Ax = b system is...

Info icon This preview shows pages 1–2. Sign up to view the full content.

Chapter 1 A x = b system is solved by Augmented Matrix Solutions from A x = 0 is like this: [ x ] = [ p ] + x 3 [ v ] Theorem 4 All of these are all true and all false at the same time 1) A x = b has solution for each b 2) b is linear combinations of col A 3) columns of A span R m 4) pivot position in every row of row reduced matrix Theorem 5 Conditions for Linear Operator 1) A ( u + v ) = A u + B v 2) A (c u ) = c A u Theorem 6 v is solution to A x = 0 p is solution to A x = b Theorem 8 If a set has more vectors than entries in each vector, then the set is linearly dependent Theorem 10 Uniqueness Matrix Each T ( x ) has a unique A such that T ( x ) = A x A is in form A = [ T ( e 1 ) T ( e 2 ) T ( e 3 ) … T ( e n )] Matrix Transformations In rotation, [1] = [cos u] A = [ cos u - sin u ] [0] [sin u] [ sin u cos u ] Onto Transformation If each b in R m is image of at least one x in R n One to One Transformation If each b in R m is image of at most one x in R n Theorem 11 T is one to one iff T ( x ) = 0 has only trivial solution Theorem 12 A) T maps R m onto R n iff columns of A span R n B) T is one to one iff col A is linearly independent Chapter 2 Theorem 3 (A + B) T = A T + B T (AB) T = B T A T x = A -1 b is solution Theorem 6 (A -1 ) -1 = A (AB) -1 = B -1 A -1 (A T ) -1 = (A -1 ) T Elementary Matrix A single elementary row operation on I Elementary row operation performed on A is EA, where E is made by doing same operation on I E is invertible, inverse of E is elementary matrix that makes E back to I Theorem 7 Matrix A is invertible iff A is row equivalent to I, and elementary row operations that reduces A to I also makes I into A -1 Invertible Matrix Theorem All of these are all true and all false at the same time A) row equivalence to I B) n pivots C) A x = 0 has only trivial solution D) linearly independent columns E) A x maps R n onto R n F) A x = b has at least one solution for each b G) columns span R n H) A x is one to one I) there exists C such that CA = I J) there exists D such that AD = I K) A T is invertible L) columns form basis of R n M) Col A = R n N) dim Col A = rank = n O) Nul A = { 0 } P) dim Nul A = 0 Q) det A != 0 R) number 0 is not eigenvalue of A Homogenous Coordinates Action Operator with 1 added to bottom left and vectors has 1 tagged on Translation vector occurs on left column Subspace Contains { 0 } vector and is close under both multiplication by scalars and addition by vectors Nullspace Set Nul A of all solutions to A x = 0 Theorem 12 Nul A is subspace of A Theorem 13 Pivot column forms basis for Colspace [ x ] B Coordinate Vector Dimension Number of vectors in basis Rank Dimension of Colspace Theorem 14 rank A + dim Nul A = number of columns Theorem 15 Basis Theorem Any linearly independent set of p elements in H where dim H = p is basis for H Chapter 3 Theorem 1 det A = a
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern