I have the matrix \begin{bmatrix}1&0&0\\2&2&-1\\0&1&0\end{bmatrix} I know that the only eigenvalue is 1 with multiplicity 3

I solved for the first eigenvalue and got \begin{bmatrix}0\\1\\1\end{bmatrix}

How do I find the other two? I know they are \begin{bmatrix}0\\1\\0\end{bmatrix} and \begin{bmatrix}1/2\\0\\0\end{bmatrix} but when I do $(A-\lambda I)v_2 = v_1$, I get the system of equations $2x + y -z = 1$, $y -z =1$. I don't see how that gives the second eigenvector.


  • 2
    $\begingroup$ Have you heard of generalized eigenvectors and methods of finding those? Your particular example requires two generalized eigenvectors and there are many such examples on MSE. For example, cfm.brown.edu/people/sg/classnotes1.pdf $\endgroup$ – Moo Apr 21 '17 at 12:01
  • $\begingroup$ What makes you think there are $3$ independent eigenvectors ? If there were, then $A$ would be diagonalizable, with only eigenvalue $1$, which would mean $A$ would be unity matrix, which it is definitely not :-) Your system for kernel of $A-\lambda I$ has two equations, so the eigenspace should have dimension $1$. $\endgroup$ – Nicolas FRANCOIS Apr 21 '17 at 12:18
  • $\begingroup$ Here is an example where algebraic multiplicity and geometric multiplicity are different. Those are important to know at least probably later course. $\endgroup$ – mathreadler Apr 21 '17 at 12:30

All you need is just to solve the system. The system of equations $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=1 \end{cases} $$ has the solution $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 0\\1+t\\t \end{bmatrix},\quad t\in\Bbb R. $$ Now take $t=0$.

The second vector is obtained similarly: solve $$ \begin{cases} 2x+y-z=1,\\ \qquad y-z=0 \end{cases} $$ to get $$ \begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 1/2\\-t\\t \end{bmatrix},\quad t\in\Bbb R $$ and set $t=0$.

P.S. Those other two vectors are not eigenvectors. They are called generalized eigenvectors.


If you use a Linear Algebra tool like in MATLAB, or a libraries for programming languages, like NumPy with Python or MathNet Numerics with C#, you can find out that the eigenvectors calculated by these tools are:

[0, 0, 0] [(√2)/2, (√2)/2, (√2)/2] [(√2)/2, (√2)/2, (√2)/2]

The eigenvalues are as stated:

(1,0) (1,0) (1,0)

1 of multiplicity 3.

We should bear in mind that eigenvalues could be complex numbers.

  • $\begingroup$ Please use MathJax for mathematical equations and expressions: math.meta.stackexchange.com/questions/5020 $\endgroup$ – jvdhooft Jun 27 '17 at 21:23
  • $\begingroup$ @jvdhooft: Thanks a lot for the advice on MathJax, will do! $\endgroup$ – GEN Jun 27 '17 at 21:42

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.