Let
$$A= \begin{pmatrix}1&-4&-1&2\\ 1&-4&1&2\\ 4&-16&-7&7\\ -1&4&8&-1\end{pmatrix} $$
Find Basis of $\mathrm{Col}(A) \cap \mathrm{Row}(A)$
I found basis of \mathrm{Row}(A) and \mathrm{Col}(A) :
$B_{\mathrm{Col}(A)}={\{(-2,0,0,-2),(0,-2,0,-4),(0,0,-2,2)}\}$
$B_{\mathrm{Row}(A)}={\{(1,-4,-1,2),(0,0,2,0),(0,0,0,-1)}\}$
and I know that $\dim(\mathrm{Col}(A) \cap \mathrm{Row}(A))=2$
any hint how to find the basis ?
Name basis vectors of $\mathrm{Col}(A)$ and $\mathrm{Row}(A)$ as $x_1, x_2, x_3$ and $y_1, y_2, y_3$. If some vector $z\in \mathrm{Col}(A)\cap \mathrm{Row}(a)$, then, simultaneously $z\in \mathrm{Col}(A)$ and $z\in \mathrm{Row}(a)$, which means, for some $\alpha_1,\alpha_2,\alpha_3, \beta_1, \beta_2, \beta_3$ $$\begin{cases} \alpha_1x_1+\alpha_2x_2+\alpha_3x_3=z \\ \beta_1y_1+\beta_2y_2+\beta_3y_3=z \\ \end{cases} \iff \alpha_1x_1+\alpha_2x_2+\alpha_3x_3 = \beta_1y_1+\beta_2y_2+\beta_3y_3 $$ Since $z\in \Bbb R^4$, it's really a system of 4 equations and 6 unknown variables, and after you solve it, you get an expression like $z=\gamma_1w_1+\gamma_2w_2$, where $w_1$ and $w_2$ are some linear combinations of $x_1, x_2, x_3, y_1, y_2, y_3$ and form the basis of $\mathrm{Col}(A)\cap \mathrm{Row}(a)$.
