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

invertible.

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

its basis.