Results 1 to 3 of 3

Math Help - characteristic polynomial and invertible matrices

  1. #1
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1

    characteristic polynomial and invertible matrices

    Given A, an n \times n matrix, with the following characteristic polynomial

    f(t) = (-1)^n t^n +a_{n-1}t^{n-1}+ \ \dotso \ + a_1t+a_0 show that A is invertible if only if a_0\neq 0

    what I have is f(t) can be written as det(A-t \cdot I) where I is the identity matrix, so I would have:

    det \left( \begin{bmatrix}<br />
0 & 0 & \dotso & 0 & -a_0 \\<br />
1 & 0 & \dotso & 0 & -a_1\\<br />
0 & 1 & \dotso & 0 & -a_2\\<br />
\vdots & \vdots & \ddots & \vdots & \vdots\\<br />
0 & 0 & \dotso & 1 & -a_{n-1}<br />
\end{bmatrix} - t\times \begin{bmatrix}<br />
1 & 0 & \dotso & 0 & 0 \\<br />
0 & 1 & \dotso & 0 & 0\\<br />
0 & 0 & \dotso & 0 & 0\\<br />
\vdots & \vdots & \ddots & \vdots & \vdots\\<br />
0 & 0 & \dotso & 0 & 1<br />
\end{bmatrix}\right)

    now from the definition of invertible there must exist a matrix B which is n \times n matrix such that A\cdot B = I

    but how do I find such a matrix, and what is the connections with a_0, beyond the fact that it can't be 0?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by lllll View Post
    Given A, an n \times n matrix, with the following characteristic polynomial

    f(t) = (-1)^n t^n +a_{n-1}t^{n-1}+ \ \dotso \ + a_1t+a_0 show that A is invertible if only if a_0\neq 0

    what I have is f(t) can be written as det(A-t \cdot I) where I is the identity matrix, so I would have:

    det \left( \begin{bmatrix}<br />
0 & 0 & \dotso & 0 & -a_0 \\<br />
1 & 0 & \dotso & 0 & -a_1\\<br />
0 & 1 & \dotso & 0 & -a_2\\<br />
\vdots & \vdots & \ddots & \vdots & \vdots\\<br />
0 & 0 & \dotso & 1 & -a_{n-1}<br />
\end{bmatrix} - t\times \begin{bmatrix}<br />
1 & 0 & \dotso & 0 & 0 \\<br />
0 & 1 & \dotso & 0 & 0\\<br />
0 & 0 & \dotso & 0 & 0\\<br />
\vdots & \vdots & \ddots & \vdots & \vdots\\<br />
0 & 0 & \dotso & 0 & 1<br />
\end{bmatrix}\right)

    now from the definition of invertible there must exist a matrix B which is n \times n matrix such that A\cdot B = I

    but how do I find such a matrix, and what is the connections with a_0, beyond the fact that it can't be 0?
    The value of a_0 is the product of the eigenvalues.
    Since it is non-zero it means the eigenvalues are not zero.
    Thus, the matrix must be invertible.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1314
    I would have pointed out that every matrix is either diagonalizable, with its eigenvalues on the main diagonal, or has a Jordan form, again with its eigenvaues on the main diagonal, both of which have determinant equal to the products of the eigenvalues. Since the determinant is the product of the eigenvalues, a matrix is invertible if and only if it does not have 0 as an eigenvalue: if a_0 is non-zero. But ThePerfectHacker's method is much better.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. invertible matrices
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 5th 2010, 11:48 PM
  2. Replies: 1
    Last Post: December 15th 2009, 07:26 AM
  3. Invertible Matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 20th 2009, 10:39 PM
  4. Invertible Matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 15th 2009, 11:19 PM
  5. invertible matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: July 21st 2005, 01:32 PM

Search Tags


/mathhelpforum @mathhelpforum