How are eigenvectors/eigenvalues of a matrix related with its invertibility? How are eigenvectors/eigenvalues of a matrix related with its invertibility? If $1,1,2$ are eigenvalues of a matrix $A$, is $A$ invertible?
 A: Yes.
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.
Edit.
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
A: Hint: For any $\lambda \in \mathbb C$, we know by definition that $\lambda$ is an eigenvalue of $A$ iff $A - \lambda I$ is not invertible.  Can you see which value of $\lambda$ is most relevant to this question?
A: The eigenvalues of a square matrix tell you how much the matrix stretches vectors in different directions. The matrix has a nontrivial nullspace when there are vectors it collapses, i.e., stretches by a factor of $0$.
