TT* + I is invertible I've the following exercise which I can't solve:
Prove that: 
$$ AA^* + I $$
is invertible for all Matrix $ A $ in finite-dimensional field $V$ with inner product. $  A^* $ is the adjoint operator. 
Any help will be appreciated. 
 A: $<AA^*v+v,v>=<AA^*v,v>+<v,v>=<A^*v,A^*v>+<v,v>$
this is true from the property of inner product and definition of $A^*$.
Let's call $A^*v$ in another name, let's call it the vector $\alpha$.
So 
$<AA^*v+v,v>=<A^*v,A^*v>+<v,v>=<\alpha,\alpha>+<v,v>=\|\alpha\|+\|v\|$
if $v \neq 0$, then this term is larger than zero. Which means $AA^*+I$ is positive definite, which means it only has positive eigenvalues, which means 0 isn't an eigenvalue, which means it is invertible.
A: Assume that there's a $v\in V$ such that $AA^*v=-v$. Can you use the definition of the adjoint to conclude that $v=0$?
A: If $ A^*A + I $ is not invertible then $ det(A^*A + I)=0 $ then $-1$ is an eigenvalue of $ A^*A $ then there exist nonzero $v$ such that $A^*Av=-v$ therefore $$ v^*A^*Av=-v^*v$$
which means nonsense because norm is not negative.
Then $ A^*A + I $ is always invertible.
Why does $v$ always imply column-vector and $v^*$ always imply row-vector?
A: Assuming that there is an inner product in $V$. Then for every $x \in V, x \neq 0$, we have 
$$x^*(AA^*+I)x = \underbrace{(A^*x)^*(A^*x)}_{\geq 0}+\underbrace{x^*x}_{>0}$$
So $AA^*+I$ is positive definite and thus invertible.
