# 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?

• Hint: the determinant of a matrix is the product of its eigenvalues. – user3294068 May 7 '14 at 14:43

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

• The thing is, we haven't done determinants in class yet. So if there is a proof without the use of determinants, it would be highly appreciated. – user95319 May 7 '14 at 15:16
• You should mention it is true because $\mathbb C$ is an algebraically closed field. – Gabriel Romon May 7 '14 at 15:16
• @user95319 I edited my answer to do that :-) – Ant May 7 '14 at 15:24

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?

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 $0$.