# Can full rank matrix have zero determinant?

all.

Let A be a real symmetric $(N\times N)$ matrix.

Although I would like to check its rank and determinant in order to calculate the inverse of A, a confliction arised. $\\$ Since A is a large matrix, (I wish I could break the matrix in several small pieces and have a look), I checked the rank and the determinant through MATLAB.

Though the rank is N, the determinant is equal to 0.

>> rank(A)
ans = N

>> det(A)
ans = 0

>> cond(A)
ans = 5.2e+05


As far as I know that the full rank is identical to be invertable, but it cannot since the determinant is zero. How it can be resolved? Thank you in advance.

• For an $n\times n$ matrix $A$, the condition that $A$ has rank $n$ is equivalent to $\det A\ne0$. Another equivalent condition is that $A$ is invertible. Apparently there are rounding problems. Jul 11 '17 at 16:42
• The large condition number suggests that your matrix is very close to a matrix that does not have full rank (which makes it appear that A has full rank but also zero determinant). Jul 11 '17 at 16:42
• If your matrix coefficients are small, even though the matrix has full rank, the determinant can underflow. (Scaling the elements by $s$ scales the determinant by $s^n$.
– user65203
Jul 11 '17 at 16:57
• It is possible that $\det$ underflows, but the matrix is still well conditioned. See scicomp.stackexchange.com/a/1330. Jul 11 '17 at 17:03
• @copper.hat Thank you! I'll have a look! The question seems quite similar to my case. Jul 11 '17 at 17:17