Proving that removing any vector of the linearly dependent set gives a linearly independent set Consider the matrix representing 6 linearly dependent vectors:
$$\left(\begin{array}{llllll}
1 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 1
\end{array}\right)$$
I know how to prove that the vectors in this matrix are linearly dependent, but how can I show (concisely) that removing any one of the vectors we get a linearly independent set?

Note that the given matrix is in its Row Reduced Echelon Form and has rank 5. That is, its columns span $\mathbb R^5$.And basis of $\mathbb R^5$ can only have 5 vectors. There can not be more than 5 linearly independent vectors in $\mathbb R^5$. Now, coming to the given matrix, remove any column (say 3rd column), then you have $5$ by 5 matrix (rearrange the columns) in row reduced form with 5 pivots, hence the columns of this matrix are linearly independent. Same conclusion if you remove any other column from the given matrix. Hence, removing any column from the matrix (given in your post) results in a full rank square matrix.
