A step in finding Eigen vectors

Question

I think a fastest way to determine an Eigen vector of a matrix corresponding to an Eigen value is by using Cramer's rule. But if two equations are identical, then applying Cramer's rule is not possible.

Edit: For example, if matrix $A=\begin{pmatrix}-1 & -1 & -1\\-2&-2&-2\\0&0&0\end{pmatrix}$. Then the system of equations is:

$x+y+z=0$

$x+y+z=0$.

$0x+0y+0z=0$. Then by Cramer's rule,

$\frac{x}{1-1}=-\frac{y}{1-1}=\frac{z}{1-1}=k$(say)

$\implies \frac{x}{0}=-\frac{y}{0}=\frac{z}{0}=k$, which gives an Eigen vector $(0, 0, 0)$, which is false (because an Eigen vector is non zero by its definition).

How to solve the above system of equations in order to determine an Eigen vector?

Answer 1

Cramer's rule is one way (and not a good way) to solve some systems of linear equations. It has little to do with finding eigenvalues and eigenvectors.

Cramer's rule fails in this case because that system is not always solvable. The determinant (the denominator in Cramer's rule) is $0$. In this example the vectors $(0,0,1)$ maps to the zero vector, so $(0,0,1)$ is an eigenvector with eigenvalue $0$. Each of $(1,0,0)$ and $(0,1,0)$ is also an eigenvector. Can you find the corresponding eigenvalues?

In general, you find eigenvalues by finding the roots of the characteristic polynomial. In this case you can do that by inspection - because the rows of the matrix are constants.

Answer 2

What makes you think that Cramer's rule is the fastest way to find eigenvectors? You give an example where it doesn't work - haven't you noticed that it doesn't work in any example?

Say $\lambda$ is an eigenvalue of $A$. To find a basis for the corresponding eigenspace we need to find the general solution to $$(A-\lambda I)x=0.$$But if $\lambda$ is an eigenvalue of $A$ then $A-\lambda I$ has determinant zero, which is precisely when Cramer's rule cannot be applied.

(In your post you don't seem to let $\det=0$ bother you. This is scary bad - you can't divide by zero!!!! Read what Cramer's rule actually says: if a certain determinant is non-zero then...)