1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra (Meaning of Rank)

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data

    True or False:
    If A is an n x n matrix, then the rank of A equals the number of linearly independent row vectors in A.

    2. Relevant equations


    3. The attempt at a solution

    Okay, I know this is a ridiculously easy question, but I'm wondering if there is a catch to it. My inclination is to say this is true. But wouldn't this be true for ANY matrix? Why does the question limit it to an n x n matrix? Are they just being tricky?
  2. jcsd
  3. Jul 15, 2009 #2


    User Avatar

    Staff: Mentor

    Well, if it were n x m with n not equal to m, then the row rank and column rank would not generally be the same, right?

    Also, if you have more rows than columns, what can you say about the linear independence of the rows...?

    http://en.wikipedia.org/wiki/Rank_(linear_algebra [Broken])

    Last edited by a moderator: May 4, 2017
  4. Jul 15, 2009 #3
    Thanks for your reply. In the wikipedia article that you cited (which I had read before posting this question) it says in the third sentence: "Since the column rank and the row rank are always equal, they are simply called the rank of A". I believe this statement is correct, which disagrees with your first point. Am I missing something?
    Last edited by a moderator: May 4, 2017
  5. Jul 15, 2009 #4


    User Avatar

    Staff: Mentor

    No, that sentence was my point. if n=m, then you use the term "rank" instead of differentiating between rows and columns. I think your original answer is correct, just that it cannot be extended always to cases where n is not equal to m. Hope I'm not just adding confusion here...
  6. Jul 15, 2009 #5
    Okay, that makes perfect sense. Thank you!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook