# Prove that if $\mathrm{rank}(A) < n$ then $\det(A) = 0$?

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?

• I guess you can use the multilinear and alternating properties to deduce the actions of the elementary row operations on determinants and then proceed to row reduce the matrix. But that's essentially your first idea, so would you consider that cheating?
– EuYu
Apr 19 '13 at 0:34
• Could you possibly start me off? Apr 19 '13 at 0:34
• I suppose that may be the point of it. So how would I begin from the fact that $\rank(A)$ < $n$? Apr 19 '13 at 0:35
• You would have to use the fact that $\rank(A) < n$ implies that the Reduced Row Echelon Form has a zero row. It's pretty much exactly how you planned to proceed. Now you're simply showing that elementary row operations preserve the "zero-ness" of the determinant, i.e. the determinant will be zero after an elementary row operation if and only if it was zero to begin with.
– EuYu
Apr 19 '13 at 0:38
• I see, that's very helpful, thank you. Apr 19 '13 at 0:43

$f(x+cy,y,z,\cdots) = f(x,y,z,\cdots) + cf(y,y,z,\cdots)=f(x,y,z,\cdots)$ 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.

Here goes the sketch of a proof.

Let $A$ be your matrix. Do Gaussian elimination on $A$ so that you end up with an upper triangular matrix $U$. During this process, you either

1) Switch rows, which switches the sign of the determinant.

2) Add a multiple of a row to another row, which preserves the determinant.

Thus, $\det(U) = \pm\det(A)$. Hence if $\det(A)\neq 0$, then $\det(U)\neq 0$. Since $U$ is upper triangular, this means that the diagonal of $U$ does not contain any zeroes.

Hence $U^{-1}$ exists and doing the 'reverse' of the Gaussian elimination done for $A$ gives $A^{-1}$ from $U^{-1}$.

You do need to work out some messy details, but I'm sure it can be made to work.

• What do you mean? Apr 19 '13 at 0:29
• Yes but I've been specifically told to use the multilinear and alternating properties of the determinant? Apr 19 '13 at 0:30
• O, sorry. I mixed some stuff up, let me see...
– Abel
Apr 19 '13 at 0:32
• I'm sure they aren't sufficient by themselves but I need to incorporate those properties into my proof... Apr 19 '13 at 0:32
• I guess you could formalise the procedure of Gaussian elimination...
– Abel
Apr 19 '13 at 0:34

Without bothering too much about the mechanics of finding echelon forms, you may reason as follows. I will suppose you know the determinant is multilinear and alternating in the columns. (You didn't specify rows or columns; in fact the determinant is multilinear and alternating both as function of the rows and as function of the columns, but the two properties are not the same.)

If you consider successive columns, and determine the rank of the matrix up to there, then there must be some $j$ where adding column number $j$ to the matrix does not increase the rank. Then that column is a linear combination of the previous columns. Now write out that column as that linear combination, apply the operation $\det$ to the matrix, and use it's multilinear property with respect to column $j$, giving a linear combination of determinants, in each of which column $j$ is a copy of some previous column $j'$. But the alternating property, all these determinants are $0$, and therefore the original determinant is.