A basic question on determinant and rank of a matrix How to prove that if the determinant of a $n \times n$ matrix is zero then the rank is less than $n$. I can prove the converse. Only a hint is enough.
My definition of rank is the maximum number of linearly dependent columns.
 A: An instructive method is row elimination. If the determinant of $A$ is zero, then our reduced echelon form matrix $R$ must have determinant zero, so it must have a zero row. ($R$ must have determinant zero because $R = E_1\dotsb E_nAF_1^T\dotsb F_n^T$, where $E_i$ and $F_i$ are elementary matrices.)
The presence of a zero row shows that the rows are linearly dependent. Row elimination also shows that the row rank of a square matrix equals the column rank, so we're done.
Please correct me if I've made a mistake with this proof; I'm just learning this myself.
A: Supposing you know (only) that $\det(A\cdot B)=\det(A)\det(B)$, you might reason as follows. We show the contrapositive: assuming the rank of the matrix is$~n$, we shall show that the determinant cannot be$~0$.
The fact that $A$ has rank$~n$ implies that $A$ has an inverse matrix$~B$ (you seem to know this, although an earlier comment said you didn't). Then $A\cdot B=I$ so $1=\det(I)=\det(A\cdot B)=\det(A)\det(B)$, and $\det(A)\neq0$.
To see that rank$~n$ implies the existence of an inverse, first observe that the column space of$~A$ is all of $\Bbb R^n$, so in particular each standard basis vector $\mathbf e_j$ lies in the column space. Putting the coefficients that express $\mathbf e_j$ as linear combination of the columns of$~A$ into a column vector $\mathbf b_j$, this expression is given by the matrix equation $\mathbf e_j= A\cdot\mathbf b_j$. Now combine these columns $\mathbf b_j$ for $j=1,2,\ldots,n$ into a matrix $B$, for which one then has $I=A\cdot B$.
A: One geometrical intuition behind this is as follows:
The determinant gives the volume of the Parallelepiped formed by the matrix rows. If this volume is zero, this means that the rows are not linearly independent and therefore that the matrix rank is smaller than $n$.
For a proof, assume that the determinant is zero. Use Gaussian elimination to get a diagonal matrix. Now, the determinant is just the product of diagonal elements, so one of them must be zero, making the rank of the matrix less than $n$.
