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Math Help - Linear Transformation

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    Linear Transformation

    Suppose T is a linear transformation on V. Show that T is invertible iff there is a unique isometry S in L(V) such that T = S(root(T*T)). I'm assuming T is the adjoint transformation to T here. Thanks in advance for any help.
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    Quote Originally Posted by joeyjoejoe View Post
    Suppose T is a linear transformation on V. Show that T is invertible iff there is a unique isometry S in L(V) such that T = S(root(T*T)). I'm assuming T is the adjoint transformation to T here. Thanks in advance for any help.
    I'm assuming that V is a finite-dimensional vector space with an inner product \langle.\,,\,.\rangle, so that the adjoint of T is given by \langle T^*x,y\rangle = \langle x,Ty\rangle\ (x,y\in V). The operator R = \sqrt{T^*T} is selfadjoint, and is invertible if and only if T is (reason: take determinants in the equation R^2=T^*T). Also, for each x\in V,

    \|Rx\|^2 = \langle Rx,Rx\rangle = \langle R^2x,x\rangle = \langle T^*Tx,x\rangle = \langle Tx,Tx\rangle = \|Tx\|^2.\qquad(*)

    So there is a (unique) isometric linear map S_0 from the range of R to the range of T defined by S_0(Rx) = Tx. If T and R are invertible then these ranges are the whole of V, and S_0 is the unique isometry on V such that T=S_0R.

    Now suppose that T is not invertible. Then S_0 is still an isometry, but neither its domain nor its range are the whole of V. However, since the ranges of T and R have the same dimension, so do their orthogonal complements. Therefore there exists a (nonzero) isometry S_1 from (R(V))^\perp to (T(V))^\perp. The operators S_+ = S_0+S_1 and S_- = S_0-S_1 are then distinct isometries on the whole of V, and they both have the property that S_\pm R = T. So in this case, S is not unique.
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