It is clear that given a matrix A, if there is a non-zero vector v such that Av=0, then A is not invertible. However, if a matrix A has determinant zero, how would you show that there exists a vector v such that Av=0 and v is not 0?
[Math] Why does a singular matrix have a (non-zero) eigenvector
linear algebra
Best Answer
If the matrix is invertible, then you can use gaussian elimination to transform it to an upper triangular one which has nonzero determinant; hence the original matrix had nonzero determinant as well. So, if the determinant is zero, the matrix is singular. You can easily show that a singular matrix has linearly dependent columns, so some $\sum_j A_{\cdot, j}\, x_j=0$, $A_{\cdot, j}$ denoting the $j$-the column of the matrix. So, for $x=(x_1,\ldots, x_n)^T$, you get that $Ax=0$.