Let $\mathcal{M}$ be the set of $5\times 5$ real matrices of rank $3$. Given a matrix in $\mathcal{M}$, the set of columns of $A$ has $2^5-1=31$ nonempty subsets. Let $k_A$ be the number of these subsets that are linearly independent.

Determine the maximum and minimum values of $k_A$, as $A$ varies over $\mathcal{M}$.
[i]The rank of  a matrix is the dimension of the span of its columns.[/i]
Given a matrix \( A \in \mathcal{M} \), where \(\mathcal{M}\) is the set of \(5 \times 5\) real matrices of rank 3, we need to determine the maximum and minimum values of \( k_A \), the number of linearly independent subsets of the columns of \( A \).

1. **Lower Bound:**
   - Since the rank of \( A \) is 3, there exist 3 linearly independent columns in \( A \).
   - Any non-empty subset of these 3 columns will be linearly independent.
   - The number of non-empty subsets of 3 columns is \( 2^3 - 1 = 7 \).
   - Therefore, \( k_A \geq 7 \).

2. **Upper Bound:**
   - Any subset of columns with more than 3 columns will be linearly dependent because the rank of \( A \) is 3.
   - We need to count the number of linearly independent subsets of columns with at most 3 columns.
   - The number of ways to choose 1 column out of 5 is \( \binom{5}{1} \).
   - The number of ways to choose 2 columns out of 5 is \( \binom{5}{2} \).
   - The number of ways to choose 3 columns out of 5 is \( \binom{5}{3} \).
   - Therefore, the total number of linearly independent subsets is:
     \[
     k_A \leq \binom{5}{1} + \binom{5}{2} + \binom{5}{3}
     \]
     \[
     k_A \leq 5 + 10 + 10 = 25
     \]

3. **Conclusion:**
   - The minimum value of \( k_A \) is 7.
   - The maximum value of \( k_A \) is 25.

The final answer is \(\boxed{7 \leq k_A \leq 25}\)