If an n x n matix A is not invertible, then is 0 an eigenvalue of A? I know that if $0$ is an eigenvalue of a matrix $A$, then $A$ is not invertible, but is the opposite also true? If the matrix isn't invertible, is the eigenvalue always $0$?
 A: Yes, if $A$ is not invertible, then $0$ is an eigenvalue of $A$. Put another way, there must be a nonzero vector $y$ that satisfies $Ay=0$ hence $y$ is an eigenvector of $A$ with eigenvalue $0$.
Indeed, $A$ not invertible $\implies$ there are two distinct vectors $y_1,y_2$ such that $Ay_1=Ay_2$ $\implies$ $A(y_1-y_2)=0$ $\implies$ $y=y_1-y_2$ is a nonzero eigenvector of $A$ with eigenvalue $0$ [since $Ay=0$ with $y =y_1-y_2 \not = 0$].
Alternate proof: Let $a_1, \ldots, a_n$ be the $n$ row of $A$ [with $A$ a square matrix]. Then $A$ not invertible $\implies$ the rows of $A$ are linearly dependent $\implies$ there are scalars $c_1,c_2,\ldots, c_n$, at least one of the $c_i$s nonzero, such that $\sum_{i=1}^n c_ia_i = 0$ $\implies$ [letting $c$ be the following vector $c= [c_1,c_2, \ldots, c_n]^{\top}$] that $Ac = \sum_{i=1}^n c_ia_i = 0$ $\implies$ $c$ is an eigenvector of $A$, and is not $0$ [because one of the $c_i$s is not $0$] with eigenvalue $0$.
A: If $A$ is not invertible, then $\det(A) = 0$.
One might find the eigenvalues of $A$ by finding the roots of its characteristic polynomial:
$$\det(A-\lambda I) = 0.$$
And since $\det(A) = 0$, $\lambda = 0$ is a root and so is an eigenvalue.
A: Yes.
One way to think about it is if a square matrix is not invertible, it is linearly dependent. If the columns of a matrix are linearly dependent, there exists a linear combinations of the columns that sums to the $0$ vector.
$A \vec{b} = \vec0$
In this situation A is acting like a scalar multiple of $\vec{b}$. This is the definition of an eigenvalue. The factor it is scaling $\vec{b}$ by, $0$, is the eigenvalue.
$A\times0 = \vec0$
Thus, $0$ is an eigenvalue for all non-invertible square matrices.
A: The condition of non-invertibility amounts to $\det(A)=0$ which in turn is equivalent to that the rank of $A$ is not maximal. But that means that you can find a non-zero column $X$ such that
$$
A\cdot X=0.
$$
Thinking of $X$ as a vector, the displayed formula says that $X$ is an eigenvector for the eigenvalue $0$.
