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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.