Showing there exists a vector u such that $T^*T(u) = T^*(v)$? Suppose $T \in L(U, V )$, and $v \in V$. Show that there exists a $u_0 \in U$ such that $(T^*T)(u_0) = T^*(v)$, and $||T(u_0) − v|| \leq ||T(u) − v||$ for all $u ∈ U$. Furthermore, show that if $\dim\operatorname{range} T = \dim U$, then $u_0 = (T^*T)^{-1}(T^*(v))$. These are finite dimensional inner product spaces and $T^*$ is the adjoint. 
So I know that $T^*(v)$ must map somewhere in $U$, but I feel like it is too simple to just define $v=T(u_0)$. Any help much appreciated!    
 A: Right, so I finally got the answer. I actually take the orthogonal projection of $v$ onto the $rangeT$ (because even if $v$ is already in $rangeT$, its projection is just itself). So we call this projection $v_0$.Then there does exist a $u_0$ such that $T(u_0)=v_0$. Then $T^*T(u_0)=T^*(v_0)$ and $ T^*(v_0)=T^*(v)$ because if $v=v_0 +w$ where $v_0$ is in $rangeT$ and $w$ would be in $ kerT^*$, then $T^*T(u_0)=T^*(v)$. Also there is an orthogonal projection property which states that $||P_v(u)||≤||u||$. So it follows that $||T(u_0)-v||≤||T(u)-v||$. 
Then, if $dim rangeT = dim U$, $T^*T$ is invertible. So $$u_0=(T^*T)^{(-1)}(T^*(v))$$ can be $$T^*Tu_0=(T^*(v))$$ so that is rather obvious how that works out.
A: Of cause, you cannot define $v=T^*u_0$ because $v$ is given and $u_0$ is unknown. Think of $T^*u_0=v$ as of a linear system to find $u_0$, but as we know not all linear systems are solvable...
You can proceed in this way: let's try to use what you already know
\begin{align}
& a) \dim\operatorname{range}T^*T=\dim\operatorname{range}T,\\
& b) \dim\operatorname{range}T=\dim U\quad\Rightarrow\quad T^*T\text{ is invertible}.
\end{align}
First we notice that the part b) is exactly your question "Furthermore, show that..." since $T^*Tu_0=T^*v$ implies $u_0=(T^*T)^{-1}T^*v$ if the matrix is invertible, so let us look at proving that $u_0$ exists.


*

*We have (by definition of range) the trivial inclusion $\operatorname{range}TT^*\color{red}\subset\operatorname{range}T$ which together with part a) gives
$$
\dim\operatorname{range}T^*T=\dim\operatorname{range}T\color{red}\ge \dim\operatorname{range}TT^*.
$$
Since the inequality is true for all $T$ we can replace $T$ with $T^*$ to get the opposite inequaity, that is, in fact, it holds
$$
\dim\operatorname{range}T^*T=\dim\operatorname{range}TT^*\stackrel{a)}{=}\dim\operatorname{range}T^*.
$$

*Now we look at what we have: by definition $\operatorname{range}T^*T\subset\operatorname{range}T^*$ and also we have proved that $\dim\operatorname{range}T^*T=\dim\operatorname{range}T^*$. The only possibility for a subspace to have the same dimension as the whole space is for them to be equal, that is
$$
\operatorname{range}T^*T=\operatorname{range}T^*.
$$
Now it is clear that $u_0$ exists. Indeed, 
$$
T^*v\in\operatorname{range}T^*=\operatorname{range}T^*T\quad\Rightarrow\quad\exists u_0\colon T^*Tu_0=T^*v.
$$

*To prove inequality: for any $u$ we have
\begin{align}
\|Tu-v\|^2&=\|Tu_0-v+T(u-u_0)\|^2=\|Tu_0-v\|^2+2\operatorname{Re}\langle Tu_0-v,T(u-u_0)\rangle+\|T(u-u_0)\|^2=\\
&=\|Tu_0-v\|^2+2\operatorname{Re}\langle \underbrace{T^*(Tu_0-v)}_{=0},u-u_0\rangle+\underbrace{\|T(u-u_0)\|^2}_{\ge 0}\ge\|Tu_0-v\|^2.
\end{align}

