Why does $\det A=0$ imply that there exists a non-zero solution to the homogeneous linear equations determined by $A$?

Question

Why does $\det A=0$ imply that there exists a non-zero solution to the homogeneous linear equations determined by $A$?

Moreover, is it an "if and only if" case? That is, if there's a non-zero solution to the set of linear equations, does the matrix determined by the coefficients automatically have a determinant of zero?

Answer 1

Having $\det A=0$ means the linear equations are not independent, hence there are more unknowns than equations, and you can choose arbitrarily a number of these unknowns – in particular, you can choose them non-zero.

Conversely, a set of non-zero solutions can be interpreted as the coefficients of a non-trivial relation between the columns of the matrix, which implies its determinant is $0$.

Answer 2

If $\det A\ne 0$, we can quite explicitly write down the invere matrix $A^{-1}$. Then $Av=0$ implies $v=A^{-1}Av=A^{-1}0=0$.

And if there is no non-zero solution to $Av=0$ then the $n$ (assuming $A$ is $n\times n$) vectors $v_i:=Ae_i$, $1\le i\le n$, with $(e_1,\ldots,e_n)$ the standard basis are linearly independent, hence a basis of $n$-dimensional space, hence $v_i\mapsto e_i$ defines a linear map and the corresponding matrix $B$ is obviously inverse to $A$ (i.e., $AB$ and $BA$ both correspond to the identity map). Then $1=\det(AB)=\det(A)\det(B)$, hence $\det(A)\ne0$.

Answer 3

For each of the three elementary row operations, applying it either does nothing to the determinant (in the case of a $R_1 \mapsto R_1 + 6 R_3$ sort of operation) or it multiplies by a non-zero scalar ($-1$ for a flip-flop, and whatever scalar you're multiplying the row by for a scalar multiplication).

So: elementary row operations can change the determinant of a matrix, but they can't change whether that determinant is zero or not.

Therefore, given a square matrix with $\det A = 0$, when we echellonize it, we must get a matrix with some $0$s on the diagonal (because a diagonal matrix with no $0$s on the diagonal doesn't have determinant $0$). It's then obvious that the corresponding homogeneous equation has a non-zero solution (corresponding to the row of $0$s in the matrix).

Conversely, if there exists a non-zero solution, then our matrix echellonizes to a matrix with a row full of zeroes, so the determinant of that matrix is zero; hence the determinant of our original matrix was zero.

Answer 4

$0=\det(A)=\det(A-0\cdot I)$, hence $0$ is an eigenvalue of $A$.