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Thread: determinant of T linear operator

  1. #1
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    determinant of T linear operator

    Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

    i) 1) Prove that det(T tIv) = det ( [T]β tI ) for any scalar t and any ordered basis β
    ii) 2) Prove that if det (T) ≠ 0 , then T is invertible
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mystic View Post
    Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

    i) 1) Prove that det(T tIv) = det ( [T]β tI ) for any scalar t and any ordered basis β
    ii) 2) Prove that if det (T) ≠ 0 , then T is invertible
    I'm not quite sure what you mean for the first one. For the second I assume you're allowed to use that the invertibility of a matrix is equivalent to invertible determinant. If so, then do the following. Let $\displaystyle V$ be an $\displaystyle F$-space. Then, there exists an isomorphism $\displaystyle \phi:V\to F^{\dim_F V}$ afforded by the choice of the ordered basis $\displaystyle \]beta$. Suppose then that $\displaystyle \det\left([T]_\beta\right)\ne 0$ then for every $\displaystyle v\in V$ one has that $\displaystyle \phi(v)\in F^{\dim_F V}$ and since $\displaystyle [T]_\beta$ is invertible one has that $\displaystyle [T]_\beta^{-1}$ exists. Prove then that $\displaystyle T\left(\phi^{-1}\left([T]_{\beta}^{-1}(\phi(v))\right)=v$. Thus, $\displaystyle T$ is surjective and the conclusion follows since $\displaystyle \dim_F V<\infty$
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  3. #3
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    it seems to me the first problem boils down to showing that [tIv]β = I, the matrix whose entries are the kronecker delta, no matter what β is, since if so, you can subtract the matrices on the RHS. you may need to invoke the isomorphism between Mn(F) and the elements of End(V) (you need β to exhibit this isomorphism, but the fact that they are isomorphic doesn't depend on β).
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    The problem defines $\displaystyle \det(T)=\det([T_{\beta}])$ . In order to see if $\displaystyle \det (T)$ is well defined we need to prove that the right side does not depends on the chosen basis. Indeed, if $\displaystyle \beta'$ is another basis of $\displaystyle V$ then, $\displaystyle [T]_{\beta}$ and $\displaystyle [T]_{\beta'}$ are similar matrices so, $\displaystyle \det([T]_{\beta})=\det([T]_{\beta'})$ .

    Now, fixed $\displaystyle \beta$ an using the standard isomorphism between $\displaystyle \textrm{End}_{\mathbb{K}}(V)$ and $\displaystyle \mathbb{K}^{n\times n}$ we have $\displaystyle [T-tI_V]_{\beta}=[T]_{\beta}-tI_n$ .
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