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

  • $\begingroup$ Note: $\lambda_2=-1$ $\endgroup$
    – EPS
    May 2, 2015 at 0:12
  • $\begingroup$ typo at beginning - fixed now - answer still same though $\endgroup$ May 2, 2015 at 0:15
  • $\begingroup$ You are confused because you are using $v_2$ for a vector as well as for a scalar at the same time! $\endgroup$
    – EPS
    May 2, 2015 at 0:16
  • $\begingroup$ i know i dont know why he has done that - but no I am confused as to how to work out, say $c_2$ instead if this is the case, i.e $-1c_1+c_2=0$ and $-1c_1+c_2=0$ $\endgroup$ May 2, 2015 at 0:18

1 Answer 1


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 $(0,1)$.

  • $\begingroup$ Why $(0,1)$? Im just confused about the $1$ to be honest, because when I look at it, it could also be 0 surely? $\endgroup$ May 2, 2015 at 0:19
  • 1
    $\begingroup$ @Bernard.Mathews For any $v_2$, the vector $(0,v_2)$ is a multiple of $(0,1)$. $\endgroup$
    – 5xum
    May 2, 2015 at 0:21
  • 1
    $\begingroup$ @Bernard.Mathews No, not just a positive integer. $(0,-0.89498345703984)$ is also an eigenvector. $\endgroup$
    – 5xum
    May 2, 2015 at 0:22
  • 1
    $\begingroup$ @Bernard.Mathews The way I understand it is that really, matrices don't have eigenvectors, they have eigenspaces. For each eigenvalue $\lambda$, the matrix $A$ has an eigenspace which is equal to all vectors for which $Ax=\lambda x$. This space is the nullspace of $Ax - \lambda x$, and it may have more than one dimension. In your particular case, the nullspace is one-dimensional, and the basis vector of the nullspace is $(0,1)$. So, in a way, when you say "the eigenvector is $(0,1)$", you are saying "the basis of the nullspace is $\{(0,1)\}$" $\endgroup$
    – 5xum
    May 2, 2015 at 0:28
  • 1
    $\begingroup$ @Bernard.Mathews And no, it's not stupid. It's a very important concept and it's vital that you fully grasp it. $\endgroup$
    – 5xum
    May 2, 2015 at 0:28

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