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Math Help - eigenvalues and invertible matrix

  1. #1
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    eigenvalues and invertible matrix

    Hello I'm having some issues with this current problem and I'm hoping that someone can help.

    The problem states: Given A is an n x n invertible matrix such that lamda is equal to 1 is not an eigenvalue, I is the identity matrix, (A^1) is the inverse matrix of A. Explain why

    a) (I-A^-1) is invertible

    b) (A^-1)(I-A^-1) = (A-I)^-1

    c) ((I-A^-1)^-1)(A^-1) = (A-I)^-1
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigenvalues and invertible matrix

    Quote Originally Posted by nivek0078 View Post
    The problem states: Given A is an n x n invertible matrix such that lamda is equal to 1 is not an eigenvalue, I is the identity matrix, (A^1) is the inverse matrix of A. Explain why a) (I-A^-1) is invertible
    \displaystyle\begin{aligned}I-A^{-1}\mbox{ is not invertible }&\Rightarrow \det (I-A^{-1})=0\\&\Rightarrow\det (A^{-1}-1I)=0\\&\Rightarrow 1\mbox{ is eigenvalue of }A^{-1}\\&\Rightarrow 1^{-1}=1\mbox{ is eigenvalue of }(A^{-1})^{-1}=A\mbox{ (Absurd).} \end{aligned}

    b) (A^-1)(I-A^-1) = (A-I)^-1
    That is false. Choose for example A=\mbox{diag }(2,3).
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  3. #3
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    Re: eigenvalues and invertible matrix

    A matrix is invertible if and only if 0 is not an eigenvalue. For any eigenvalue, \lambda, there exist some non-zero vector such that Av= \lambda v. In that case, A^{-1}Av= v= \lambda A^{-1}v and so A^{-1}v= \frac{1}{\lambda} v. In other words, \lambda is an eigenvalue of A if and only if \frac{1}{\lambda} is an eigenvalue of A^{-1}, with the same eigenvectors. In particular, if 1 is not an eigenvalue of A, 1/1= 1 is not an eigenvalue of A^{-1}. (The fact that A is invertible means it does not have 0 as an eigenvalue so that reciprocal always exists.)

    Then (I- A^{-1})v= v- A^{-1}v= v- \frac{1}{\lambda}v= (1- \frac{1}{\lambda})v. The eigenvalues of I- A^{-1} are of the form 1- \frac{1}{\lambda}. The fact that 1 is not an eigenvalue of A means 1 is not an eigenvalue of A^{-1} and so 1- 1= 0 is not an eigenvalue of I- A^{-1} which means that I- A^{-1} is invertible.

    Unless I have made a silly computation error (always possible) (b) and (c) are NOT true! As a counter-example, look at A= \begin{pmatrix}2 & 1 \\ 0 & 2\end{pmatrix}.
    Thanks from nivek0078
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