{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture_5 - MA 35100 LECTURE NOTES FRIDAY JANUARY 22 Rank...

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

View Full Document Right Arrow Icon
MA 35100 LECTURE NOTES: FRIDAY, JANUARY 22 Rank of a Matrix Say we have a system of equations. Denote A as its augmented matrix. We may compute the reduced row-echelon form E = rref( A ), and use it to solve this system by considering the pivots. We solve for the leading variables in terms of the free variables. Of course, we can rearrange the equations so that sometimes we can express the free variables in terms of the leading variables. Ultimately what matters is the number of leading variables, not really which leading variables we have. We use this to make a definition. Definition. The rank of a matrix A is the number of pivots in rref( A ). We denote this nonnegative integer by rank( A ). We give a brief example. Consider the following matrix: A = 1 2 3 4 5 6 7 8 9 . We compute that the reduced row-echelon form for this matrix is E = rref( A ) = 1 0 - 1 0 1 2 0 0 0 . Note that this matrix has two pivots (which are circled above), so that the rank is rank( A ) = 2. We note that we have defined the rank of a matrix , not the rank of a system of equations. This definition is a but more subtle, so we defer more discussion until a later lecture.
Image of page 1

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

View Full Document Right Arrow Icon
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