From $EA=R=\operatorname{RREF}(A)$, recover the four subspaces. Nonzero rows of $R$ form a row-space basis. Solving $Rx=0$ yields a right-null-space basis. Pivot-column indices of $R$ select a basis from the original columns of $A$. Bottom rows of $E$ give left annihilators; over $\mathbb R$ transpose them to obtain $\ker A^T$, and over $\mathbb C$ conjugate-transpose them to obtain $\ker A^*$.
For numerical problems, orthogonal factorizations or SVD can be more stable than exact-style RREF calculations.
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