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Thread: need to proof

  1. #1
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    need to proof

    Hi, I need to prove this statement, but my teacher wants me to so without using adj{A} and I kind of stuck because using Cramer's rule to prove that is almost like using the adj{A}.
    " A\in M_n(\mathbb{Q}) therefore A\in M_n(\mathbb{R})
    prove that: if A is nonsingular in M_n(\mathbb{R}), A is also nonsingular in M_n(\mathbb{Q})"
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  2. #2
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    Re: need to proof

    Can you show that A^{-1} has only rational entries?
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  3. #3
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    Re: need to proof

    Quote Originally Posted by xbox360 View Post
    Hi, I need to prove this statement, but my teacher wants me to so without using adj{A} and I kind of stuck because using Cramer's rule to pr ove that is almost like using the adj{A}.
    " A\in M_n(\mathbb{Q}) therefore A\in M_n(\mathbb{R})
    prove that: if A is nonsingular in M_n(\mathbb{R}), A is also nonsingular in M_n(\mathbb{Q})"
    I assume that $M_n(\mathbb{R})$ is the set of all $n\times n$ matrices having real entries. If that is correct then $M_n(\mathbb{Q}) \subset M_n(\mathbb{R})$. Any nonsingular square matrix has non-zero determinate. Can you explain what you are and are not allowed to do for a proof.
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  4. #4
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    Re: need to proof

    One way to do this is to transform the problem to vector spaces. Here's the relevant theorem:

    Let V be a vector space of finite dimension n over a field F and T a linear transformation from V to V. Then $T^{-1}$ exists if and only if Ker(T)=0. If $T^{-1}$ exists, $T^{-1}$ is a linear transformation.

    You can use this theorem to prove your statement (without determinants at all). If you don't see how to do this or need help on the theorem, post again.
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  5. #5
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    Re: need to proof

    Quote Originally Posted by xbox360 View Post
    Hi, I need to prove this statement, but my teacher wants me to so without using adj{A} and I kind of stuck because using Cramer's rule to prove that is almost like using the adj{A}.
    " A\in M_n(\mathbb{Q}) therefore A\in M_n(\mathbb{R})
    prove that: if A is nonsingular in M_n(\mathbb{R}), A is also nonsingular in M_n(\mathbb{Q})"
    @xbox360, why do you not answer the replies to your post? Truth be told, what exactly is the question?

    Is it A is nonsingular in M_n(\mathbb{R}) then A is also nonsingular in M_n(\mathbb{Q})~?

    Or is it A\in M_n(\mathbb{Q}) and is nonsingular in M_n(\mathbb{R}), then A is also nonsingular in M_n(\mathbb{Q})~?

    Please tell us.
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