I have the following proposition to prove:
Proposition (1) Let $A:=\{\vec{v}_1,...,\vec{v}_n\}$ be a subset of an $F$-vectorial space $V$. If $\text{span}(A)=V$ then every subset of $V$ independient in $V$ has a cardinality less than or equal to the cardinality of $A$.
This is the proof I found:
Proof. Suppouse exists a subset $B:=\{\vec{b}_1,...,\vec{b}_m\}$ of $V$ independient in $V$ such that $\text{card}(B)>\text{card}(A)$, I mean, $m>n$. $\text{span}(A)=V$ if and only if every vector in $V$ can be expressed as a linear combination of vectors in $A$, so $$\mathbf{A}:=\begin{bmatrix} a_{11} & ... & a_{1n}\\ \vdots & \vdots & \vdots\\ a_{n1} & ... & a_{nn} \end{bmatrix}\begin{bmatrix} \vec{v}_1\\ \vdots\\ \vec{v}_n \end{bmatrix}=\begin{bmatrix} \vec{e}_1\\ \vdots\\ \vec{e}_n \end{bmatrix}$$ has solution for all $\vec{e}_1,...,\vec{e}_n\in V$ and in consequence $\mathbf{A}$ is invertible. Let $\mathbf{A}^{-1}=\begin{bmatrix} c_{11} & ... & c_{1n}\\ \vdots & \vdots & \vdots\\ c_{n1} & ... & c_{nn} \end{bmatrix}$. Then $$\begin{bmatrix} \vec{v}_1\\ \vdots\\ \vec{v}_n \end{bmatrix}=\begin{bmatrix} c_{11} & ... & c_{1n}\\ \vdots & \vdots & \vdots\\ c_{n1} & ... & c_{nn} \end{bmatrix}\begin{bmatrix} \vec{b}_1\\ \vdots\\ \vec{b}_n \end{bmatrix}$$ so each vector in $A$ can be expressed as a linear combination of $\vec{b}_1,...,\vec{b}_n$. But we have that $\vec{b}_{n+1}$ can be expressed as a linear combination of vector of $A$, and each ocurrence of vector of $A$ in such linear combination can be replaced by a linear combination of $\vec{b}_1,...,\vec{b}_n$, then $\vec{b}_{n+1}$ can be expressed as a linear combination of $\vec{b}_1,...,\vec{b}_n$, but this implies that $B$ is not independiente in $V$, and this is absurd. $\blacksquare$
I want to know if is this proof is right, can anyone help me? If I am wrong please don't tell me where I am wrong, just tell me I am wrong. Thank you so much.