I have a question where I am trying to find the general solution of a linearised system, which I have linearised. I am just having difficulty obtaining the correct corresponding eigenvectors to my eigenvalues. I am usually fine at these but I am not getting the correct answer (I have solutions). I will highlight my issue below.
My matrix A for my linearised system is given by
$$A=\begin{bmatrix}-1 & 0 \\-1 & 0\end{bmatrix}$$
and my eigenvalues are $\lambda_1=0$ and $\lambda_2=-1$ which I have found with no problem
However, I then know eigenvectors $v_1$ and $v_2$ are found as follows, again usually fine with:
$(A-\lambda_1 I)v_1=0$ and $(A-\lambda_2 I)v_2=0$
i.e $(A-\lambda_1 I)v_1= \left( \begin{array}{ccc} 0 & 0 \\ -1 & 1 \end{array} \right)$$v_1=0$
So, we have $\left( \begin{array}{ccc} 0 & 0 \\ -1 & 1 \end{array} \right)\left( \begin{array}{ccc} v_1 \\ v_2 \end{array} \right)=0$
So $0v_1+0v_2=0$ and $-1v_1+1v_2=0$
i.e $-1v_1=-1v_2 \implies v_1=v_2=1$ so corresponding eigenvector to eigenvalue $\lambda_2=-1$ is $\left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)=0$
This is all fine. Here is where I am stuck. Now to find the corresponding eigenvector to eigenvalue $\lambda_2=0$
We want to find $v_2$ such that $(A-\lambda_1 I)v_2= Av_2= \left( \begin{array}{ccc} -1 & 0 \\ -1 & 0 \end{array} \right)$$v_2=0$
So this is where I am stuck:
trying to find eigenvector $v_2$ we have $-1v_1+0v_2=0$ and $-1v_1+0v_2=0$
The solution says $v_2=\left( \begin{array}{ccc} 0 \\ 1 \end{array} \right)$ However I cannot see how the writer has got this. Please forgive my stupidity but obviously if $-1v_1+0v_2=0$ then $v_1=0$ but how do they find $v_2$? the fact that both $v_2$ coefficients are zero is confusing me, any help much appreciated! Thanks
$\endgroup$ 41 Answer
$\begingroup$First of all you have confusing notation, as you use $v_2$ for both the vector and the component of the vector.
That said, in my answer, $v$ is a vector with components $v_1$ and $v_2$
You have a system of equations, the first $$-v_1= 0$$
and the second $$-v_1 =0$$
Now, any pair $v_1, v_2$ which satisfies both equations is an eigenvector. Clearly, $v_1$ must be $0$, and clearly, if $v_1=0$, the pair will satisfy the system for ANY value $v_2$. Therefore,
$$(A-0\cdot I) v = 0$$
if and only if $v$ is a multiple of $(0,1)$.
$\endgroup$ 14
