1. ## 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.

2. Originally Posted by joeyjoejoe
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.