Let's say that I have a function in two variables: $f(x,y)$ and I know that $(x_{0},y_{0})$ is the point of minimum for $f$.
Now imagine I do a generic linear transformation on the coordinates:
$\begin{pmatrix} x\\ y\end{pmatrix}\rightarrow \begin{pmatrix} X\\ Y\end{pmatrix}=A \cdot \begin{pmatrix} x\\ y\end{pmatrix}$
where $A$ is a 2 by 2 matrix.
My question is : is it true that the point minimum of $f(X,Y)$ (call it $(X_{0},Y_{0})$) is the transformed of the old point of minimum (by the same transformation $A$)? So:
$\begin{pmatrix} X_{0}\\Y_{0} \end{pmatrix}=A \cdot \begin{pmatrix} x_{0}\\y_{0} \end{pmatrix} $
Is this true or false in general? And in the second case, what are the properties that $A$ needs to have in order for it to be valid?
I ask this because I wanted to determine the point of min of a function by first doing a linear transformation on the coordinates, such that it is easier to find it, and then applying on the point the inverse transformation to find it in the original old variables.
Thanks everyone!
As you asked in the comments whether there was a specific theorem for your initial claim, and I answered "I don't think so", here is a proof.
We'll have functions of one variable only, because having two variables does not change anything here: the unique variable can have a domain which is the cartesian product of the two variables domains.
$x_0$ is a minimum for $f$ on its domain $D$, is equivalent to:
$\forall x \in D, f(x) \ge f(x_0)$.
We do a change of variables: $X = A(x)$, where $A$ is injective on $D$.
Define $g$ on $A(D)$:
$\forall X \in A(D), \exists !\,x \in D, A(x)=X$ because $A$ is injective, so we can define $g(X)=f(x)$.
Then $\forall x \in D, g(A(x))=f(x)$.
Let $X_0=A(x_0)$. Then
$\forall X \in A(D), \exists !\,x \in D, X=A(x), f(x) \ge f(x_0),$
$g(X)=g(A(x))=f(x) \ge f(x_0)=g(A(x_0))=g(X_0)$.
We have proven $\forall X \in A(D), g(X) \ge g(X_0)$: $X_0$ is a minimum of $g$ on $A(D)$.
$A$ being injective, $A$ is a bijection between $D$ and $A(D)$, so $A^{-1}$ is well defined. Then if you find $X_0$ minimum of $g$ on $A(D)$, $x_0$ is obtained by $x_0=A(X_0)$.
Note: we have reasoned for "a" minimum. If $x_0$ is "the" minimum of $f$ on $D$ - meaning there is no other minimum - the reasoning is the same, except that
$\forall x \in D, f(x) \ge f(x_0)$ shall be replaced by
$\forall x \in D, x \ne x_0 \Rightarrow f(x) \gt f(x_0)$.