If $A$ is an $n \times n$ matrix with $\DeclareMathOperator{\rank}{rank}$ $\rank(A) < n$, then I need to show that $\det(A) = 0$.
Now I understand why this is – if $\rank(A) < n$ then when converted to reduced row echelon form, there will be a row/column of zeroes, thus $\det(A) = 0$
However, I have been told to use the fact that the determinant is multilinear and alternating and subsequently deduce that if $\det(A)$ is non-zero, $A$ is invertible.
How do I use the properties of the determinant to prove these claims?
Best Answer
$f(x+cy,y,z,\cdots) = f(x,y,z,\cdots) + cf(y,y,z,\cdots)=f(x,y,z,\cdots)$ using multilinearity and the alternating property respectively.
Hence you can add columns together without changing the determinant. But the rank is $<n$ if and only if some linear combination of the columns is trivial, in which case we can obtain an equivalent determinant with a column of zeros. By linearity, the determinant is zero.