TITLE: How to prove that if the determinant of the matrix is zero then at least one eigenvalue must be zero?
QUESTION [20 upvotes]: For a matrix $A$, if $\det(A)=0,$ prove and provide an example that at least one eigenvalue must be zero.

At first, I tried using the identity that the product of eigenvalues is the determinant of the matrix, so it follows that at least one must be zero for the determinant to be zero. Is this correct? Could I also prove it by using $(A-\lambda I)X=0$, for some $X\neq 0?$
If $\lambda=0,$ then we have $AX=0$, but I can't say $\det(A)\cdot \det(X)=0$ because $X$ is not a square matrix and doesn't have a determinant. How would I continue?

REPLY [16 votes]: I do not know what you know about the determinant and how you think of it, but the determinant of a square matrix $A$ is zero iff the matrix is not invertible, and that is equivalent to the kernel being non-trivial, which means that $Ax=0$ for some $x\ne0$.