Answer
See the explanation below.
Work Step by Step
a. The equality $dim Row A + dim Nul A = n$
follows from the rank-nullity theorem.
The $rank of A$, which is the dimension of the row space of $A$, is equal to the number of linearly independent rows of $A$.
The nullity of $A$, which is the dimension of the null space of $A$, is equal to the number of free variables in the row-reduced echelon form of $A$.
Therefore, $dim Row A + dim Nul A$ is equal to the total number of rows in the row-reduced echelon form of $A$, which is equal to $n$.
Rearranging this equality gives $dim Row A = n - dim Nul A$.
b. The columns of $A$ are linearly independent if and only if the rows of $A^T$ are linearly independent.
Therefore, the $rank of A$ is equal to the rank of$ A^T$.
Applying the rank-nullity theorem to $A^T$, we have;
$dim Col A^T + dim Nul A^T = m$,
where $dim Col A^T$ is the dimension of the column space of $A^T$ and $dim Nul A^T$ is the dimension of the null space of $A^T$.
Since the rank of A is equal to the $rank of A^T, dim Col A = dim Col A^T$. Therefore, $dim Col A = m - dim Nul A^T = m - dim Nul A$, where the last equality follows from the fact that the null space of $A^T$ is the same as the null space of A.
