# Eigenvalue Problem and Singularity

I am interested in one detail regarding the calculation of eigenvalues and eigenvectors. The equation one starts with is for a matrix A, a vector x and a scaler $\lambda$ the following:

Ax = $\lambda$x

where x $\neq$ 0. How does it follow now, that the matrix A-I*$\lambda$ is singular (i.e. its determinant is equal to zero)?

-
add comment

## 2 Answers

Since $Ax=\lambda x$ we have $(A-\lambda I)x=0$. Since $x\neq 0$ and $x\in\ker(A-\lambda I)$, we have $\ker(A-\lambda I)\neq\{0\}$. In particular, $\det(A-\lambda I)=0$.

-
Alright, since the kernel does not only contain {0}, the matrix is not injective, hence surely not bijective and thus singular, correct? –  GuestGuest Mar 3 '13 at 21:57
add comment

$$\lambda \text{ eigenvalue of } A \\ \iff \text{exists nonzero vector }v\text{ with }Av = \lambda v\\ \iff \text{exists nonzero vector }v\text{ with }(A - \lambda I)v = 0\\\iff \text{exists nonzero vector }v\in\ker(A-\lambda I)\\ \iff A - \lambda I\text{ singular}$$

-
add comment