Consider T ∈ L(U, V ). U, V are finite dimensional. Show that dim range (T*T) = dim range T. Then, show that if dim range T = dim U then T*T is invertible. T* is the adjoint of T.
So it is obvious that T*T: U -> U but I just have no clue where to start or show this (or both parts rather). I was thinking of starting with a basis for both U and V but I'm not sure. Any help would be much appreciated!
I'm assuming these are inner product spaces, otherwise you need to define the adjoint map differently (in a general normed vector space, the adjoint is a map between dual spaces $T^*: V^* \to U^*$).
We show that their kernels are equal first. Note that is $x \in \text{ker}(T)$ then $$Tx = 0 \,\,\,\,\,\, \implies \,\,\,\,\,\, T^*Tx =0$$ so $x \in \text{ker}(T^*T)$. Conversely, if $x \in \text{ker}(T^*T)$, then $$T^*Tx = 0 \,\,\,\,\, \implies \,\,\,\,\, \langle x, T^*T x\rangle = 0 \,\,\,\,\, \implies \,\,\,\,\, \langle Tx, Tx \rangle = 0 \,\,\,\,\, \implies Tx = 0.$$ Thus $x \in \text{ker}(T)$. Hence $\text{ker}(T) = \text{ker}(T^*T)$. Now by the rank nullity theorem we have $$\text{dim}(\text{range}(T)) + \text{dim}(\text{ker}(T))= \text{dim}(U)$$ and $$\text{dim}(\text{range}(T^*T)) + \text{dim}(\text{ker}(T^*T))= \text{dim}(U).$$ Subtracting these two equations and using that the kernels have the same dimension yields the result.
