Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue. I can't find the missing link between singularity and zero eigenvalues as is stated in the following proposition:
A matrix $A$ is singular if and only if $0$ is an eigenvalue.
Could anyone shed some light?

If 0 is an eigenvalue, then there exists a vector $v$ in your space such that $A.v = 0$. If your matrix size is 4x4 with one 0 eigenvalue and you write the image of the eigenvectors, you get:
$$(v11, v12, v13, 0)$$
$$(v21, v22, v23, 0)$$
$$(v31, v32, v33, 0)$$
$$(v41, v42, v43, 0)$$
You can see it's singular because:


*

*the 3 vectors cannot possibly span a 4-dimensional space, it's an hyperplane, a 3 dimensional sub-space in 4 dimensions (so it would be a line in two dimension, a plane in 3 dimension)

*any point not on the hyperplane cannot be described as a combination of the 3 column vectors described above. So not all points can be reached through multiplication by $A$.

*so the transformation that is $A$ cannot always be inverted, because for all $y$ there is not always a point $x$ such that $Ax=y$  (for example, the $y$ not on the hyperplane)

*so $A$ is singular, it cannot be inverted in general


hopes this rolls out the reasoning clearly enough.
