If the row echelon form of a square matrix has no zero row, it is invertible. Otherwise, it is singular.
Why?
If the row echelon form has a zero row, in a linear system, it has either no solution or infinitely many solutions. So, is invertibility linked to having only one solution?
Is there a geometrical interpretation for my question?
Best Answer
Excellent. Your request for a geometric interpretation shows me that you are on the right track in learning linear algebra! (Well, at least visualizing the standard 1-3 dimensions)
Consider the Reduced Row Echelon Form (RREF) of a matrix
A
, it concisely describes some of the subspace information associated withA
.The RREF tell us:
The null space being trivial (i.e, consisting of only an appropriate null vector) implies that the column space of
A
occupies the entirety of it's dimension(equal to the column count of A) and that there is no linear combination of any vectors in it's range that reduce to 0 vector.The process of matrix inversion is supposed to find a subspace which when multiplied with
A
gets projected to the appropriate identity matrix.If there is any linear combination of columns of
A
that reduces to 0, then it cannot be reversed to map onto it's original linear combination, which means that the vector is nullified. (Mapped to the 0 vector). This is exactly what the rank of a matrix succinctly describes with mathematical beauty.The exact same can be witnessed and verified on the column space and null space of the non-invertible matrix
A'
(which are incidentally, NOT coincidentally, the row space and left-null space ofA
)