I have been trying to find the eigen vectors of this matrix. Actually i have already found the eigen values.

$A=\begin{pmatrix} 1&-2&0\\ -2&5&0\\ 0&0&2\\ \end{pmatrix}$

where, I have found the eigenvalues

$\lambda_1=5.83,\lambda_2=2, \lambda_3=0.17$

but i have no idea to find the eigenvectors, although i have tried to solve the equation.

My answer is different from what it should be.

The correct answer for the eigenvectors should be : $(0.383, -0.924,0)^T$, $(0,0,1)^T$, and $(0.924, 0.383,0]$ where my answer for the first and the third eigenvectors yield to $(0,0,0)^T$

I don't have any idea to solve it. Any help will be appreciated, thanks!

  • $\begingroup$ An eigenvector is by definition distinct from $(0,0,0)^T$. $\endgroup$
    – Crostul
    Commented Jan 7, 2017 at 11:19
  • $\begingroup$ This may be of help, it describes a simple case: sosmath.com/matrix/eigen2/eigen2.html $\endgroup$
    – NoChance
    Commented Jan 7, 2017 at 11:20

2 Answers 2


Just solve the equations $(A-\lambda_iI)v=0$. This is a system of rank $2$, hence the subspace of solutions (i.e. the eigenspace) has dimension $1$.

B.t.w., the non-integer roots are $3\pm2\sqrt2$, not $5.83$ and $0.17$ (these values are only approximations).


For the eigenvalue $\lambda_1=3+2\sqrt2$, one has to solve the linear system: $$(A-\lambda_1I)v=\begin{bmatrix}-2(1+\sqrt 2)&-2&0\\-2&2(1-\sqrt 2)&0\\0&0&1-2\sqrt2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Proceed with row reduction: it is easy to see that row 2= row 1×$(\sqrt2-1)$, hence the system is equivalent to the system deined by the matrix: $$\begin{bmatrix}-2(1+\sqrt 2)&-2&0\\0&0&0\\0&0&1-2\sqrt2\end{bmatrix}\rightsquigarrow\begin{bmatrix}-2(1+\sqrt 2)&-2&0\\0&0&0\\0&0&1-2\sqrt2\end{bmatrix}\rightsquigarrow\begin{bmatrix}1+\sqrt2&1&0\\0&0&1\\0&0&0\end{bmatrix}.$$ The solutions satisfy $z=0$ and $(1+\sqrt 2)x+y=0$, so that $$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x\\-(1+\sqrt 2)x\\0\end{bmatrix}=x\underbrace{\begin{bmatrix}1\\-(1+\sqrt 2)\\0\end{bmatrix}}_{\text{eigenvector }e_{\lambda_1}}$$

  • $\begingroup$ what does it mean with " This is a system of rank 22, hence the subspace of solutions (i.e. the eigenspace) has dimension 11" ? Sorry, I don't understand, $\endgroup$ Commented Jan 7, 2017 at 11:30
  • $\begingroup$ The system of linear equations $(A-\lambda_iI)v=0$ is equivalent to 2 independent equations. As we're in dimension $3$, this implies the solutions are a subvector space of $\mathbf R^3$ with dimension 1.. In other words, the solutions are multiples of a single non-zero vector. $\endgroup$
    – Bernard
    Commented Jan 7, 2017 at 11:34
  • $\begingroup$ can you give one example about how to solve it ? Sorry, i still don't understand $\endgroup$ Commented Jan 7, 2017 at 11:39
  • $\begingroup$ @Kenny Karnama: I've added the example of an eigenvector for the first eigenvalue. Hope this is clearer. $\endgroup$
    – Bernard
    Commented Jan 7, 2017 at 13:08

You have two problems. One is round-off error. The matrix $A-\lambda I$ necessarily has determinant equal to zero. If you round off $\lambda$, then the resulting approximation to $A-\lambda I$ has a very small determinant, so when you solve the equation for $v$, you're essentially dividing by a tiny number, which, depending on round-off, can give wildly different answers. E.g., in one dimension solve $0.005 x = 6$ and you get $x=1200.$ But round to two decimal places and you solve $0.01 x = 6$ and so $x=600$. The situation can be much worse with matrices and also more hidden. Probably at this point in your studies, it's safest to do as Bernard says in his answer, and use the radical versions of the eigenvalues. Save decimal approximations for the end. There are techniques for "conditioning" matrices to dodge these problems, but that's a whole other course.

Second problem: The vector $(0,0,0)$ is always a solution to $(A-\lambda)v = 0$. The whole point of the eigenvector is that it's a non-zero solution to the equation. The eigenvalue $\lambda$ is exactly the value that causes the determinant of $A-\lambda I$ to be zero, which is what you need in order for the equation to have infinitely many solutions. (Which means you can find a non-zero solution.) If the determinant were not zero, then the only solution would be $(0,0,0)$ and that's not an eigenvector.


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