How do I show that $rank(A) < n \iff det(A) = 0$ where $A$ is a $n$ x $n$ matrix
For the forward relation, I have shown that $rank(A) < n \Rightarrow$ atleast one column of $A$ can be written as a linear combination of the other columns and since $rank(A) = rank(A^t) < n$ we can say the same thing about the rows of A. So we get $a_k = c_1a_1 + ... + c_na_n$ where $a_is$ are rows of $A$ and $a_k$ is some linear combination of the other rows except itself. Hence when computing $det(A)$ we can break this matrix into the sum of determinants. In each sum, we will have 2 same rows hence their determinant will be $0$ hence $det(A) = 0$
For the backward relation, we just say that $det(A) = 0 \Rightarrow A$ is not invertible, hence it is not full rank i.e. $rank(A) < n$
Are these both correct proofs?