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Math Help - matrix canonical form

  1. #1
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    matrix canonical form

    let A= \begin{bmatrix}6 & 3 \\ 1 & 6\end{bmatrix}

    be a matrix over the field
    K = F_11(the field of integers mod p). What is the rank of A, and what is its canonical form for equivalence? Briefly justify your answer.

    Rank of A is 2 because there are two linearly independant rows or columns. How do i start solving the other part? what is its canonical form for equivalence?

    Thank you.
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  2. #2
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    Quote Originally Posted by charikaar View Post
    let A= \begin{bmatrix}6 & 3 \\ 1 & 6\end{bmatrix}

    be a matrix over the field
    K = F_11(the field of integers mod p). What is the rank of A, and what is its canonical form for equivalence? Briefly justify your answer.

    Rank of A is 2 because there are two linearly independant rows or columns. How do i start solving the other part? what is its canonical form for equivalence?

    Thank you.

    Nop, it's rank cannot be two since its determinant is zero...
    If you now evaluate its characteristic polynomial you'll find the matrix has two different eigenvalues and thus its Jordan Canonical form is...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Nop, it's rank cannot be two since its determinant is zero...
    If you now evaluate its characteristic polynomial you'll find the matrix has two different eigenvalues and thus its Jordan Canonical form is...

    Tonio
    Isn't det(A)=36-3=33? Do i have to do anything with field K=F_11?

    thanks
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  4. #4
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    Quote Originally Posted by charikaar View Post
    Isn't det(A)=36-3=33? Do i have to do anything with field K=F_11?

    thanks

    Of course you have to do "anything" with the field \mathbb{F}_{11} : it is the field of definition of your matrix and thus you have to work out ALL the operations (mod 11)! Including the determinant.

    Tonio
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  5. #5
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    I know there are 11^4 matrices in and elements of are 1,2,4,8,5,10,9,7,3,6.

    Can you help me a bit further as I still can't solve the problem.

    thanks
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