Suppose T ∈ L(V). Prove that T is invertible if and only if 0 is
not a singular value of T.
I kinda have a love-hate feeling about proofs like these. I mean I have
so many ideas, but I don't know which one to use.
I have three routes for the reverse direction using a converse.
All three routes will use an orthonormal basis under sqrt(T*T)
since sqrt(T*T) is positive.
1. Show that there is at least one eigenvector in null sqrt(T*T)'s basis
and considering that dim rangeT=dim range sqrt(T*T), and both
dim range T and dim range sqrt(T*T) is <dim range V, range T is not
2. Show that if 0 is a value on sqrt(T*T)'s diagonal, 0 must also be a
value on T's diagonal.
3. State that since 0 is on sqrt(T*T)'s diagonal, and
any value on the isometry S's diagonal (when multiplied by 0) will
give 0, S(sqrt(T*T)) will map ej to a 0 vector and since Ssqrt(T*T)=T,
ej must be a vector in nullT (or null T must have at least one eigenvector in