Results 1 to 2 of 2

Math Help - Matrices A subspaces of Mn(R)

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    5

    Matrices A subspaces of Mn(R)

    I am trying to revise for my upcoming exam and I am stuck on a question-

    Let Mn(R) be the voctor space of all nxn real matrices, n>/2 (greater than or equal to). Which of the following set of matrices A in Mn(R) are subspaces of Mn(R)?
    I) all invertible A
    II) All non-invertib;e (i.e. singular) A
    III) all A such that AB=BA for some fixed matrix B in Mn(R)
    IV) all A such that A^2=A

    i have proves I) because to be a subspace it must contain the zero element and o is not invertible.

    II) Let a= (0 0 b= (1 0
    0 1) 0 0)
    a+b = (1 0
    0 1) which is the identity and invertible so A is not closed under addition so therefore not a subspace

    III) Here i let A= (a 0 and B= (c 0
    0 b) 0 d)
    then AB = (ac 0 BA= (ca 0
    0 bd) 0 db)
    then I have said it contains the identity so is not empty
    A+B = (a+c 0
    0 b+d) which is still in Mn(R) as it is diagonal
    and then x (a 0 = xA
    0 b)
    = (xa 0
    0 xb) which is also in Mn(R) as still diagonal therefore a subspace.
    However my teacher says I am wrong because I need to prove for arbitrary nxn matrices. What does she mean?

    IV) and the same for this one.

    I look forward to hearing from you and really appreciate you reading this.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,712
    Thanks
    1470
    Quote Originally Posted by katiedavies1990 View Post
    I am trying to revise for my upcoming exam and I am stuck on a question-

    Let Mn(R) be the voctor space of all nxn real matrices, n>/2 (greater than or equal to). Which of the following set of matrices A in Mn(R) are subspaces of Mn(R)?
    I) all invertible A
    II) All non-invertib;e (i.e. singular) A
    III) all A such that AB=BA for some fixed matrix B in Mn(R)
    IV) all A such that A^2=A

    i have proves I) because to be a subspace it must contain the zero element and o is not invertible.

    II) Let a= (0 0 b= (1 0
    0 1) 0 0)
    a+b = (1 0
    0 1) which is the identity and invertible so A is not closed under addition so therefore not a subspace

    III) Here i let A= (a 0 and B= (c 0
    0 b) 0 d)
    then AB = (ac 0 BA= (ca 0
    0 bd) 0 db)
    then I have said it contains the identity so is not empty
    A+B = (a+c 0
    0 b+d) which is still in Mn(R) as it is diagonal
    and then x (a 0 = xA
    0 b)
    = (xa 0
    0 xb) which is also in Mn(R) as still diagonal therefore a subspace.
    However my teacher says I am wrong because I need to prove for arbitrary nxn matrices. What does she mean?
    You are asked about the set of all n by n matrices, A, such that AB= BA for a specific B. You cannot assume that A is diagonal (which you do) and you cannot assume 2 by 2 matrices.

    1) The set must be non-empty. Most often it is best to show that the 0 vector is in the set. Is it true that 0B= B0?

    2) The set must be closed under addition. If AB= BA and CB= BC, that is, if A and B are in the set, is (A+ C)B= B(A+ C)?

    3) The set must be closed under scalar multiplication. If AB= BA, that is, if A is in the set, and p is any scalar, is (pA)B= B(pA)?

    IV) and the same for this one.

    I look forward to hearing from you and really appreciate you reading this.

    Thanks!
    A counter-example will show that a general statement is NOT true, which is why your first examples were correct. But an example cannot prove that a statement is true.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 27th 2011, 08:07 AM
  2. Replies: 2
    Last Post: November 25th 2010, 06:34 PM
  3. Total matrices and Commutative matrices in GL(r,Zn)
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: August 16th 2010, 02:11 AM
  4. Subspaces
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 26th 2010, 05:37 AM
  5. Replies: 0
    Last Post: October 13th 2009, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum