I am watching this new lecture by Gilbert Strang.
I have the following question.
Let $A = \begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}$.
Prof. Strang showed that the column rank of $A$ is equal to the row rank of $A$.
His proof is like the following:
Because $$\begin{bmatrix}3\\4\\12\end{bmatrix} = 1 \begin{bmatrix}2\\3\\5\end{bmatrix}+ 1 \begin{bmatrix}1\\1\\7\end{bmatrix},$$
the column rank of $A$ is $2$ and
$$\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}=\begin{bmatrix}2&1\\3&1\\5&7\end{bmatrix}\begin{bmatrix}1&0&1\\0&1&1\\\end{bmatrix}.$$
Then he says that $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are the basis of the row space of $A$ because $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are linearly independent and each of the vectors $\begin{bmatrix}2\\1\\3\end{bmatrix}$, $\begin{bmatrix}3\\1\\4\end{bmatrix}$, $\begin{bmatrix}5\\7\\12\end{bmatrix}$ is a linear combination of $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$.
Don't we need to show that $\begin{bmatrix}1\\0\\1\end{bmatrix}$, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ are in the row space of $A$?
For example, $$\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}=\begin{bmatrix}2&1&3\\3&1&4\\5&7&12\\\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}.$$
$\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ are linearly independent and each of the vectors $\begin{bmatrix}2\\1\\3\end{bmatrix}$, $\begin{bmatrix}3\\1\\4\end{bmatrix}$, $\begin{bmatrix}5\\7\\12\end{bmatrix}$ is a linear combination of $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$. But $\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ are not the basis of the row space of $A$.
$\begin{bmatrix}1\\0\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\\0\end{bmatrix}$, $\begin{bmatrix}0\\0\\1\end{bmatrix}$ generate $\mathbb{R}^3$ which is bigger than the row space of $A$.
We have $A=C R$ as in the lecture. the row space of $A$ is the column space of $A^T$ and $A^T=R^T C^T$ so the colums space of $A^T$ is generated by the columns of $C^T$ (using coefficients from $R^T$) and those columns are just the rows of $C$ again.