# Finding eigenvectors and linear dependence

I am referring to the example for finding eigenvectors at http://lpsa.swarthmore.edu/MtrxVibe/EigMat/MatrixEigen.html

The given matrix is $\begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}$

First eigenvalue ($\lambda_1$) is $-1$. In order to find corresponding eigenvector we do the following - $A\cdot v_1 = \lambda_1 \cdot v_1$

Where $v_1=\begin{bmatrix}v_{1,1} \\ v_{1,2}\end{bmatrix}$.

Fortunately we get following $\begin{bmatrix} 1 & 1 \\ -2 & -2 \end{bmatrix} \cdot \begin{bmatrix}v_{1,1} \\ v_{1,2}\end{bmatrix}=0$

Here second row in first matrix is a multiple of first row. This is very convenient because then only one relation can be found between $v_{1,1}$ and $v_{1,2}$.

Is it always the case that the rows of the matrix $A-\lambda \cdot I$ be multiples of each other? If not, how does one find out eigenvectors when the rows are not multiples of each other?

## 1 Answer

You know that $v_{1,1} = -v_{1,2}$. This is always the case, as your matrix is not inversible.

If the matrix were inversible (ie, rows not multiple of each other) then you would find $(0,0)$ as the only solution (ie, no eigenvector !). This is when there is a mistake in the computation of eigenvalues.