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Math Help - Invertibility Proof

  1. #1
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    Invertibility Proof

    The question is:


    Let
    A be a skew–symmetric n×n–matrix with entries in R, i.e. A^T=-A

    Prove that In+A is an invertible matrix.

    I have tried solving this both algebraically and non-algebraically. Algebraically I have started with the fact In+A=In-A, played around starting with there, including multiplying both sides by a nxn invertible matrix B, however never really get anywhere there.

    Other then that, things that I have noticed that probably are of importance is the fact that the diagonals of the matrix In+A will be all ones, however im not sure if i need to use that and if so how how.

    Any help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by joe909 View Post
    The question is:


    Let
    A be a skew–symmetric n×n–matrix with entries in R, i.e. A^T=-A

    Prove that In+A is an invertible matrix.

    I have tried solving this both algebraically and non-algebraically. Algebraically I have started with the fact In+A=In-A, played around starting with there, including multiplying both sides by a nxn invertible matrix B, however never really get anywhere there.

    Other then that, things that I have noticed that probably are of importance is the fact that the diagonals of the matrix In+A will be all ones, however im not sure if i need to use that and if so how how.

    Any help would be greatly appreciated
    If A^{\textsc t} = -A then, for x,y\in\mathbb{R}^n, \langle Ax,y\rangle = \langle x,A^{\textsc t}y\rangle = -\langle x, Ay\rangle = -\langle Ay,x\rangle. In particular, if x=y it follows that \langle Ax,x\rangle = 0.

    Now suppose that (I_n+A)x=0. Then 0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle, and hence x=0. Thus I_n+A is injective and therefore invertible.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    If A^{\textsc t} = -A then, for x,y\in\mathbb{R}^n, \langle Ax,y\rangle = \langle x,A^{\textsc t}y\rangle = -\langle x, Ay\rangle = -\langle Ay,x\rangle. In particular, if x=y it follows that \langle Ax,x\rangle = 0.

    Now suppose that (I_n+A)x=0. Then 0 = \langle (I_n+A)x,x\rangle = \langle x,x\rangle + \langle Ax,x\rangle = \langle x,x\rangle, and hence x=0. Thus I_n+A is injective and therefore invertible.
    I would like to start by thanking you for time and help.

    Im slightly confused by the explanation though, mainly because I dont think I have been taught the material that used for the proof. What exactly does the notation <x,y> represent, I have never seen that before? Thanks again.
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  4. #4
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    Quote Originally Posted by joe909 View Post
    What exactly does the notation <x,y> represent, I have never seen that before?
    It is the inner product, or dot product. Maybe you are used to the alternative notation x.y ? In the dot notation, the key thing about the adjoint is the fact that Ax\cdot y = x\cdot A^{\textsc t}y.
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