Results 1 to 3 of 3

Math Help - Verifying the diagonalizability of a 2x2 matrix with rank 1

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    22

    Verifying the diagonalizability of a 2x2 matrix with rank 1

    Hello,

    If A is a 2x2 matrix with rank 1, and one eigenvalue is known to be nonzero, then is A diagonalizable? I think this matrix is diagonalizable because if you take a generic 2x2 matrix with rank one:
    [ a b ]
    [ax bx]
    and you find the characteristic polynomial, (a-t)(bx-t) - bax = 0, then one eigenvalue must be zero and thus A is diagonalizable. Is this reasoning correct? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    \displaystyle \begin{bmatrix}<br />
a & b\\ <br />
0 & 0<br />
\end{bmatrix}

    \displaystyle p(\lambda_A)=\lambda^2-tr(A)\lambda=\lambda(\lambda-tr(A))=0

    \displaystyle \lambda_1=tr(A)=a \ \ \begin{bmatrix}<br />
0 & b\\ <br />
0 & -a<br />
\end{bmatrix}\rightarrow\begin{bmatrix}<br />
1 & 0\\ <br />
0 & 0<br />
\end{bmatrix}\rightarrow x_2\begin{bmatrix}<br />
0 \\ <br />
1 <br />
\end{bmatrix}

    \displaystyle \lambda_2=0 \ \begin{bmatrix}<br />
a & b\\ <br />
0 & 0<br />
\end{bmatrix}\rightarrow x_2\begin{bmatrix}<br />
\frac{-b}{a}\\ <br />
1 <br />
\end{bmatrix}

    It is only diagonalizable if b\neq 0.

    A matrix is only diagonalizable if it has n linearly independent eigenvectors. Also, just because a matrix has distinct eigenvalues it may not be diagonalizable if the eigenvectors aren't lin. independent, and conversely, a matrix with repeated eigenvalues isn't necessarily non-diagonalizable either.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,232
    Thanks
    1795
    If a 2 by 2 matrix has rank 1 then it has non-trivial kernel. That is, there exist non-zero vectors, v, such that Av= 0= 0v so that 0 is an eigenvalue. If it also has a non-zero eigenvalue, then it has two distinct eigenvalues, so two independent eigenvectors and is diagonalizable.

    d.w. smith, eigenvectors corresponding to distinct eigenvalues are independent so two distinct eigenvalues imply two independent eigenvectors so, for a 2 by 2 matrix, diagonalizable.

    As for your example, if b= 0, then \begin{bmatrix}a & 0 \\ 0 & 0\end{bmatrix} already is diagonal! It makes no sense to say it is "only diagonalizable if b\ne 0"/
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Test for diagonalizability of matrix and linear operator
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: January 22nd 2011, 10:22 PM
  2. Replies: 3
    Last Post: August 20th 2010, 06:32 AM
  3. Replies: 1
    Last Post: April 1st 2010, 09:34 AM
  4. Verifying that a matrix is a rotation matrix
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: October 27th 2009, 12:43 PM
  5. Rank of Matrix ??
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 8th 2008, 08:58 AM

Search Tags


/mathhelpforum @mathhelpforum