Results 1 to 5 of 5

Math Help - [SOLVED] Matrix^1000

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    [SOLVED] Matrix^1000

    A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)

    What is A^{1000}

    It's not diagonalizable.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Spec View Post
    A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)

    What is A^{1000}

    It's not diagonalizable.
    <br />
\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)^n = \left(\begin{array}{cc}1&-2n\\0&1\end{array}\right)<br /> <br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Spec View Post
    A=\left(\begin{array}{cc}1&-2\\0&1\end{array}\right)

    What is A^{1000}

    It's not diagonalizable.
    one way is to use induction to show that A^n=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}. a better way, which can be used for more complicated matrices, is to write A=I + B, where I is the identity matrix and

    B=\begin{pmatrix}0& -2 \\ 0 & 0\end{pmatrix}. then B^2=0 and I,B commute. then by the binomial theorem: A^n=(I+B)^n=I^n + nB=I + nB=\begin{pmatrix}1 & -2n \\ 0 & 1 \end{pmatrix}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    Quote Originally Posted by NonCommAlg View Post
    ...=(I+B)^n=I^n + nB=...
    I don't get that part. I know what the binomial theorem says, but I've never seen it used on matrices.

    EDIT: Nevermind, I didn't see the B^2 part.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Spec View Post

    I don't get that part. I know what the binomial theorem says, but I've never seen it used on matrices.

    EDIT: Nevermind, I didn't see the B^2 part.
    this is a good method of finding powers of matrices. as another example we want to find A^n where A=\begin{pmatrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}. so we let B=\begin{pmatrix}0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}.

    then A=I+B, \ B^2=\begin{pmatrix}0 & 0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, and B^3=0. again I,B commute and so we can use binomial theorem, which gives us:

    A^n=(I+B)^n=I + nB + \frac{n(n-1)}{2}B^2 = \begin{pmatrix}1 & na & nb + \frac{n(n-1)}{2}ac \\ 0 & 1 & nc \\ 0 & 0 & 1 \end{pmatrix}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Elementary matrix, restore to identity matrix
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: February 13th 2010, 09:04 AM
  2. [SOLVED] How do you find matrix this kind of matrix?
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 4th 2010, 04:22 AM
  3. [SOLVED] Chem 1000: Specific Heat problem
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: April 13th 2009, 09:10 AM
  4. 1-1000 how many 3's?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: February 5th 2009, 05:58 PM
  5. 1 coin tossed 1000 times vs 1000 coins
    Posted in the Advanced Statistics Forum
    Replies: 14
    Last Post: December 20th 2007, 06:55 PM

Search Tags


/mathhelpforum @mathhelpforum