How are eigenvectors/eigenvalues of a matrix related with its invertibility? If $1,1,2$ are eigenvalues of a matrix $A$, is $A$ invertible?
The determinant of a matrix is the product of its eigenvalues..if none of them is zero, then the determinant will be $\neq 0$ and thus A is invertible
On the other hand, if one of then is zero then so is the determinant and A is not invertible.
Suppose $\lambda = 0$ is eigenvalue, then there exists $ x \neq 0$ such that $Ax = \lambda x = 0$, so the kernel of A contains $x$ and thus A is not invertible.
Suppose now all the eigenvalues are $\neq 0$. Suppose also that (by absurd) there exists $x \neq 0$ such that $Ax = 0 = 0x$ . This would imply that $0$ is an eigenvalue, which is absurd; so $x=0$ and A is invertible