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.

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    $\begingroup$ What definition of rank are you using? $\endgroup$ – Mariano Suárez-Álvarez Aug 25 '13 at 3:36
  • $\begingroup$ @JonasMeyer corrected $\endgroup$ – RIchard Williams Aug 25 '13 at 3:48
  • $\begingroup$ You do not mean opposite but converse, probably. $\endgroup$ – Mariano Suárez-Álvarez Aug 25 '13 at 3:49
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    $\begingroup$ prasenjit: Do you know how to prove that if the rank is $n$ then the matrix has an inverse? $\endgroup$ – Jonas Meyer Aug 25 '13 at 4:03
  • $\begingroup$ @JonasMeyer : no $\endgroup$ – RIchard Williams Aug 25 '13 at 4:45

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 $I=A\cdot B$.

  • $\begingroup$ very nice explanation. Thanks $\endgroup$ – RIchard Williams Aug 25 '13 at 6:33

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.

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    $\begingroup$ You seem to be using that transforming a matrix to row-reduced echelon form leaves invariant (1) the determinant of a square matrix (2) the row rank of the matrix (3) the column rank of the matrix. These are all true, but I just want to make clear that you need to use (and therefore have proved) these, in order to be able to conclude about the original matrix $A$. $\endgroup$ – Marc van Leeuwen Aug 25 '13 at 5:26
  • $\begingroup$ Good input. I was perhaps a little concerned about the clarity of my claim that if a reduced row echelon matrix has determinant zero, then it must have a zero row... $\endgroup$ – Eric Auld Aug 25 '13 at 5:30

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 $n$.


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