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I have the following problem occuring in a longer proof. I think this should be easy, but I dont see this at the moment, maybe someone has an idea (or even an counterexampel).

Let ${\bf y} \in \mathbb{R}^n$ be a vector. Let ${\bf A} \in \mbox{GL}_n^+(\mathbb{R}^n)$ be a matrix with positive determinant whose first column is $\bf y$. Is then the following statement true?

$$ {\bf A}^{-1} {\bf y} = {\bf e}_1 $$

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Yes.

In general the $i$-th column of a matrix $A$ is always equal to $Ae_i$ where $e_i$ denotes the standard basis of $\mathbb R^n$.

Therefore we have $$ Ae_1 = y.$$

Since $A$ is invertible you multiply by $A^{-1}$ on both sides to get

$$ e_1 = A^{-1}y.$$

The only assumption needed is that the matrix $A$ is invertible i.e. that $\det A \neq 0$.

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