Results 1 to 12 of 12

Math Help - Prove: A^2+I=0 for A=[3x3] is imossible

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    133

    Prove: A^2+I=0 for A=[3x3] is imossible

    Hello everyone!

    I'm stuck on this linear quetion that is supposed to be easy: Show that no 3x3 matrix Aexists such that A^2+I=0.

    Since the question is asking for a 3x3 matrix I figured it needed the long and boring way, so I considered a matrix and initialized its entries to [A_1 A_2 A_3] where A_1=[a \, x \, u]^t, A_2=[b \, y \, v]^t, and A_3=[c \, z \, w]^t, carried out matrix product... LOST!

    2nd way: A^2+I=0,A^2+AA^{-1}=0 \,\text{ so }A+A^{-1}=0. But what then??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    What is your field? Remember, a matrix is a linear map between two vector spaces over a specific field...Basically, are you working over \mathbb{Q}, \mathbb{R} or \mathbb{C}?

    I am asking you this because the field does matter; let I be the identity matrix, and assume we are working over \mathbb{C}. Then (iI)^2+I=0, as (not) required.

    Thus, your field does matter so you'll need to be careful and you will need to use your field. However, your second approach looks promising. Essentially, you have that A^{-1} = -A. So, hit your matrix with a stick - stick in variables for your 9 entries and see what happens when you take its inverse or negate it (there will be a more subtle way, but I can't think of it at the moment!).

    EDIT: Note that A^{-1} exists because A^2=-I so A^4=I. So splitting I into AA^{-1} makes sense...(I'm sure you know this, but I thought I should probably point it out anyway).
    Last edited by Swlabr; July 7th 2010 at 03:03 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Can't you use determinants? If you've got A^{2}=-I, then it must be that \det^{2}(A)=-1. (We know that \det(-I)=-1, since for a diagonal matrix, the determinant is simply the product of the elements along the diagonal, and you have an odd number of -1's there.)

    EDIT: Swlabr's point about the field over which you have the vector space is, I agree, important.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2010
    Posts
    133
    I am working in the field of real numbers (never worked otherwise)...

    The thing is I'm plugging these variables, but then when carrying out the multiplication, I'm getting sums of products of two terms, e.g. a^2,ab,cv,xy,y^2,etc.
    and by equating my result to [tex]-IMATH] I am set to get like 9 equations with 24 unkowns!!! This is because I preferred working with linear-algebra than non-linear algebra (which I know nothing about)...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2010
    Posts
    133
    Ackbeet: The question I was asking has a 2nd part can be rephrased as: find a 2x2 matrix where the above relation holds.
    (W. Keith Nicholson, Linear Algebra with Applications, 3.2 #16)

    so if (det\,A)^2=-1 do you mean that there is something wrong? Suppose it is absurd in general, but HEY, I got a 2x2 matrix that satisfies the relation!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    If the determinant squared is equal to a negative number, then the determinant must be complex. Can that ever happen for a real matrix?

    For the 2 x 2 case, the determinant of -I is actually positive. Generalizing, for n x n matrix -I, the determinant is (-1)^{n}. So the even-n case and the odd-n case are fundamentally different.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2010
    Posts
    133
    Aha

    to cut the proof short, can we say: (\det\,A)^2=-1\Rightarrow \ \det A \ \text{doesn't belong to} \ R so we cannot find a real valued 3x3 matrix that has this property. i.e. does what I've said complete the proof?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    You'd need to reference a theorem somewhere that says if you have an n x n matrix over the reals, its determinant must be real. The determinant of an n x n matrix just involves "basic four" arithmetic, so the result is, at least, intuitive. Does Nicholson have a theorem like that?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2010
    Posts
    133
    Well so far he haven't mentioned complex numbers, although I see a few later on in the book.
    Well my professor says that he mentions things like span, rank and nullity very early in the book but then goes into details later on. So I guess your suggested theorem stands right now...
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by rebghb View Post
    Hello everyone!

    I'm stuck on this linear quetion that is supposed to be easy: Show that no 3x3 matrix Aexists such that A^2+I=0.

    Since the question is asking for a 3x3 matrix I figured it needed the long and boring way, so I considered a matrix and initialized its entries to [A_1 A_2 A_3] where A_1=[a \, x \, u]^t, A_2=[b \, y \, v]^t, and A_3=[c \, z \, w]^t, carried out matrix product... LOST!

    2nd way: A^2+I=0,A^2+AA^{-1}=0 \,\text{ so }A+A^{-1}=0. But what then??


    I think you also can say that the only way that A^2+I=0 is if to A^2 has -1 on diagonal and 0 on all other places, you deduce that A is not in M_3(R)
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    If you use the Laplace expansion or the Leibniz formula for the determinant, I think you'll see this result pop right out.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8
    Nevermind, I was wrong
    Last edited by Haven; July 11th 2010 at 12:01 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a(AB)=(aA)B=A(aB) ..
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 29th 2010, 05:14 AM
  2. Prove: f is one-to-one iff f is onto
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: June 25th 2010, 11:02 AM
  3. Replies: 2
    Last Post: August 28th 2009, 03:59 AM
  4. Please prove
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 7th 2009, 02:58 PM
  5. Prove this .
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: February 18th 2009, 05:09 AM

Search Tags


/mathhelpforum @mathhelpforum