True. If the columns of an $n \times n$ matrix $A$ are linearly dependent, then they do not span $\mathbb{R}^n$.

Here's an explanation:

1. Linear Dependence and Span: For a set of vectors to span $\mathbb{R}^n$, they must be linearly independent. Linear independence means that no vector in the set can be written as a linear combination of the others.

2. Matrix Columns and Rank: The rank of a matrix is the maximum number of linearly independent columns (or rows) it has. For an $n \times n$ matrix $A$ to have columns that span $\mathbb{R}^n$, the rank of $A$ must be $n$.

3. Linear Dependence Implications: If the columns of $A$ are linearly dependent, then the rank of $A$ is less than $n$. Consequently, there are not enough linearly independent vectors to span an $n$-dimensional space.

Therefore, if the columns of matrix $A$ are linearly dependent, they cannot span $\mathbb{R}^n$.

Would you like more details or have any questions?

Here are some related questions for further understanding:

1. How do you determine if a set of vectors is linearly independent?
2. What is the rank of a matrix, and how is it calculated?
3. Can a set of vectors span a space if they are not linearly independent? Why or why not?
4. What is the relationship between the determinant of a matrix and the linear independence of its columns?
5. How do the concepts of eigenvalues and eigenvectors relate to the span and linear dependence of a matrix?

Tip: Remember that the rank of a matrix provides crucial information about the linear independence of its columns and the dimensions of the space they span.