Confusion about minimizer(s) of convex functions I was thinking about convex functions and kind of got myself confused.
Let's say that our function is $f(x) = x^TAx + b^Tx$. And A is a symmetric either positive semidefinite or positive definite matrix (question on this later). And we want to minimize $f(x)$.
$\Delta f(x) = 2Ax + b$
and the Hessian is $= 2A$
Okay, so we know that if the Hessian is positive semidefinite then f is convex. Then for $\tilde x$ if $\Delta f(\tilde x) = 0$ then $\tilde x$ is our global minimizer.
Here comes the confusion part:
If A is not positive definitive but positive semidefinite then there can be some eigenvalues that are 0. This means A is not full rank. Then the nullspace is not empty.
We have an $\tilde x$ such that $\Delta f(\tilde x) = 0$. But for another $x_1$ that is in the nullspace of A, cant we add it to $\tilde x$ and get the same result? Since $Ax_1 = 0$?
Then how can we have a global minimizer if we have infinitely many solutions to the derivative?
Then I thought, maybe when the hessian is positive definitive our global minimizer is a point. But if the hessian is only positive semidefinite the global minimizer is a line, plane, etc. But I couldn't figure out the relation between the Hessian's nullspace and $\tilde x$. For instance, what happens if the Hessian is not full rank and there is no solution to $ \Delta f(x) = 0$. Can this happen? For the example above, I think it can happen if the vector $b$ is not in the column space of A. Then even though our function is convex, do we have no minimizers? How can this happen?
 A: You could see this problem as the sum of two convex function. First one is a $(N-1)$ dimension quadratic function while the other one is a $N$ dimensional affine function as
$$
f(x) = f_1(x) + f_2(x)
$$
where $f_1(x) = x^HAx$ and $f_2(x) =b^Hx $. 
Assume in a 2-dimensional case, we would have $ x = [x_1 , x_2]^T $ and $A$ as
$$
A = \left[\begin{matrix}
    \lambda_{11}       & 0  \\ 
    0       & 0  \\
\end{matrix}\right](1)
$$
On the other hand, for positive semi definite the inequality holds
$$ 
 x^HAx \geq 0
$$
for every x. If we substitute $A$ from equation (1) into the inequality:
$$
x^HAx = [x_1 , x_2]\left[\begin{matrix}
    \lambda_{11}       & 0  \\ 
    0       & 0  \\
\end{matrix}\right] \left[ \begin{matrix} x_1 \\ x_2\end{matrix}\right]  
$$
$$
= [x_1\lambda_{11} , 0]\left[ \begin{matrix} x_1 \\ x_2\end{matrix}\right]  = x_1^2\lambda_{11}
$$
As $\lambda_{11}$ is positive, the inequality holds for all the values of $x_1$. In fact, for $f_1(x)$ the variable $x_2$ does not have any influence on the output of optimization and the reason why is that there is no need for an additional subspace. One could define an arbitrary number of variables for a convex problem and yet the key parameters (the ones with non-zero $\lambda$) would be unique regardless of the answer of the other ones.
In this example, let's say the global minimizer is $x = [x_1^*,x_2^*]$ so
$$
f_1(x^*) = (x^*)^HA(x^*) = (x_1^*)^2\lambda_{11} 
$$
If we assume that there is another point $\widetilde{x} = [x_1^*,\tilde{x_2}] $:
$$
f_1(\widetilde{x}) = (x_1^*)^2\lambda_{11} = f_1(x^*)
$$
For $f_2(x)$ as it is an affine function, there is no minimum on the additional subspace. since it could be written as
$$
 b_1^*x_1^*+b_2^*\tilde{x_2} <  b_1^*x_1^*+b_2^*x_2^*
$$
$$
 \tilde{x_2} <  x_2^*
$$
which means for all the selected values of $x_2$, a lower one could be found.
It could be concluded that the values chosen in the null space of Hessian become crucial only when there exists additional subspace for affine function. (In this example when parameter $b_2$ has non-zero values)
