Prove that if the row vectors of a $n\times n$ matrix spans $\mathbb{R}^n$, then the columns are linearly independent I was able to prove that if a $n \times n$ matrix has linearly independent columns, then the row vectors span $\mathbb{R}^n$. However, I am unsure how to prove the other way. So far, my proof is:
Let the row vectors of A be $\vec r_{1},...,\vec r_{n}$. Then, $ span(\mathbb{R}^n)=\{\vec r_{1},...,\vec r_{n}\}$ and there exists some scalars $c_{1},...,c_{n}$ such that $c_{1} \vec r_{1}+...+c_{n}\vec r_{n}=\mathbb{R}^n$
I don't know if what I have done so far is correct.
Thanks in advanced for you help.
 A: if $R_i = \sum\limits_{j=1}^{n} a_{i,j} e_{i,j}$ for $ 1 \leq i \leq n$ are $n$-linearly  independents row vectors that span $\mathbb{R}^n$, then $C_j = \sum\limits_{i=1}^{n} a_{i,j} e_{i,j}$ are $n$-linearly independents column vectors that span $\mathbb{R}^n$. Let $x=\sum\limits_{i=1}^{n} \lambda_i R_i \in \mathbb{R}^n$,  we have:
\begin{align}
x &=\sum\limits_{i=1}^{n} \lambda_i R_i\\
& = \sum\limits_{i=1}^{n}\lambda_i \sum\limits_{j=1}^{n} a_{i,j} e_{i,j}\\
& = \sum\limits_{j=1}^{n}\lambda_j \left(\sum\limits_{i=1}^{n} a_{i,j} e_{i,j}\right)\\
& = \sum\limits_{j=1}^{n}\lambda_jC_j
\end{align}
We can change indexation of $\lambda_i$ to $\lambda_j$ and switch the sum signs because they are a finite sums.
A: One way to see this can be that if the rows span $\mathbb R^n$, then the determinant of the $n\times n$ matrix is non-zero.
This means determinant of the transpose of the $n\times n$ matrix is also non-zero.
This means we have $n$ columns which are linearly independent.
This means the columns span $\mathbb R^n$.
