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?
 A: Assuming that by “singular”, you mean a square matrix that is not invertible:
Lemma: If $A$ is invertible and $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.
Let $x$ be the eigenvector of $A$ corresponding to eigenvalue $\lambda$.  By definition, $Ax = \lambda x$.  Left-multiply by $A^{-1}$, giving $A^{-1}Ax = A^{-1} \lambda x$.  The LHS is equal to $x$ ($A^{-1}A = I$ by definition) and the RHS is equal to $\lambda A^{-1} x$ (because matrix × scalar commutes), so $x = \lambda A^{-1} x$.  Divide both sides by $\lambda$, giving $\frac{1}{\lambda} x = A^{-1} x$.  By definition, $\frac{1}{\lambda}$ is thus an eigenvalue of $A^{-1}$.

So, if $0$ were an eigenvalue of $A$, then $\frac{1}{0}$ would be an eigenvalue of $A^{-1}$.  But $\frac{1}{0}$ isn't a number, so $A^{-1}$ can't exist either.
A: $A$ singular $\iff\det(A)=0\iff\det(A-0\cdot I)=0\iff 0$ is eigenvalue of $A$.
Michael
A: 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.
A: I all depends on your starting definition. Here is one way, suppose $v$ is the eigen vector associated with $\lambda=0$ then $Av=0v=0$. Since $v\ne 0$ by definition then you have a nontrivial vector in the null space of $A$ that makes $A$ singular.
A: An $n \times n$ matrix, $\mathbf A$, is singular if and only if there is a non zero column vector $\mathbf x$ such that $\mathbf A \mathbf x = 0 = 0 \mathbf x$, i.e.,  $0$ is an eigenvalue.
A: Note that, the determinant of $n\times n$ matrix $A$ can be computed using the eigenvalues as

$$ |A|=\lambda_1\lambda_2\dots\lambda_n ,$$

which is the product of the eigenvalues. 
A: We know that $0 \in \lambda(A)$ iff there exists some nonzero solution to the eigenvector equation $Ax = \lambda x = 0\cdot x = 0$. Thus $0$ is an eigenvalue iff $\exists b \in \mathrm{Ker}(A)$ with $b \neq 0$. But since $\mathrm{Ker}(A) \neq \{ 0 \}$ we conclude that  $A$ must be singular.
A: very true. can take it like this: any matrix can be diagonalized by using appropriate elementary matrices and we know the eigen values of diagonal matrices are the diagonal elements and so if any of the eigen value is zero then determinant value of matrix is zero and so it is Singular.
