When does a nullspace of a matrix equal the trivial space?

Question

This may be a dumb question...
If we have a linear system Ax = 0, then wouldn't the nullspace of A always be {0} if there are no free variables in the system of equations?

and by no free variables, i mean all the rows in A are linearly independent

Answer 1

When the matrix has rank equal to dimension of x. You can find the rank of a matrix by doing row reduction. If the matrix is square and small, you can also take the determinant. 0 determinant = nontrivial nullspace (not invertible).

Edit: If all the rows are linearly independent, then yes, the nullspace must be trivial. This is a direct consequence of the definition of "linearly independent." It is of course also a special case of what I said above (if rows linearly independent then rank = number of rows and number of rows is less than or equal to number of columns, but the number of columns is the dimension of x).

Answer 2

If the rows of A form a linearly independent set of vector, then yes, I agree, the nullspace of A is {0} and the only solution is to Ax=0 is x=0.

Another way to say "the rows of A form a linearly independent set of vector" is to say that "A is invertible", or, equivalently, "the determinant of A is nonzero".

The only way the equation Ax=0 has nonzero soltions is if
"the determinant of A is zero", or, equivalently, if "the rows of A form a linearly dependent set of vector", or, equivalently, "A is not actually invertible".

Answer 3

If the rows of A form a linearly independent set of vector, then yes, I agree, the nullspace of A is {0} and the only solution to Ax=0 is x=0.

Another way to say "the rows of A form a linearly independent set of vector" is to say that "A is invertible", or, equivalently, "the determinant of A is nonzero".

The only way the equation Ax=0 has nonzero soltions is if
"the determinant of A is zero", or, equivalently, if "the rows of A form a linearly dependent set of vector", or, equivalently, "A is not invertible".