Prove that a square matrix $A$ has rank at most $k$ iff all its $(k+1) \times (k+1)$ minors have determinants zero.
I am not sure how this is true. say $A \in \mathbb R^{n \times n}$. Suppose $A$ has rank at most $k$. Then for any $(k+1) \times (k+1)$ minor, say of column $a_{i_1},...,a_{i_{k+1}}$, one of its columns must be a linear multiple of the others. Thus we also have a linear dependency among the columns in that minor, so that the determinant is zero.
Suppose that $A$ has rank at least $k+1$. Then the matrix $A$ has $k+1$ linear independent columns, say $\vec{v_1},...,\vec{v_{k+1}}.$ Then the matrix $S := [ \vec{v_1} \cdots \vec{v_{k+1}}]$ has rank $k+1$. Since column rank equals row rank and $S$ has clumn rank of $k+1$, $S$ has $k+1$ linearly independent rows, say at $i_1,...,i_{k+1}$-th rows. Then we can obtain a $(k+1) \times (k+1)$ whose determinant is nonzero in the original matrix $A$ by selecting the minor with columns $\vec{v_1},...,\vec{v_{k+1}}$ and rows $i_1,...,i_{k+1}$-th rows.
