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This is a matrix, A, I came across while studying eigenvectors from here.

\begin{bmatrix} 1 & 2 \\ 2 & 4 \\ \end{bmatrix}

where $AX=0$

I tried solving by using Gaussian elimination as suggested here. But I kept getting $y = 0, z = 0$. So then I solved it by just finding the ratio as shown here. Which gave me the correct answer of $y = 2 , z = -1$.

I understood that Gaussian failed because row transformations do not preserve Eigen values/vectors. But I'm confused as to when we can apply it and when we can't.

I'm a newbie to linear algebra, so it is possible my understanding is wrong.

Edit : This(image) is where I actually got confused. When I tried to find the Eigen vectors using Gaussian elimination, I kept getting $y = 0, z = 0$. Whereas by the ratio method I got the correct values. enter image description here

And that's why I was wondering why Gaussian elimination didn't work here and how to know when to not apply it.

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    $\begingroup$ Gaussian elimination can find solutions to linear systems, but by itself it doesn't find eigenvectors. The closest we can come is when a matrix is singular (like yours), and a nonzero solution to the homogeneous linear system for that matrix is an eigenvector corresponding to the zero eigenvalue. Your matrix has eigenvalues $0$ and $5$. $\endgroup$
    – hardmath
    May 13, 2018 at 19:35
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    $\begingroup$ Solutions to linear equations is rarely the same as solving polynomials such as the characteristic equation. This may merely be coincidence caused by the law of small numbers. en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers $\endgroup$
    – CyclotomicField
    May 13, 2018 at 19:39
  • $\begingroup$ @CyclotomicField The OP was to find eigenvectors, not eigenvalues. $\endgroup$
    – Arnaud Mortier
    May 13, 2018 at 19:42
  • $\begingroup$ @hardmath 's answer can be generalised to any eigenvalue, other than that I can see no way to use Gaussian elimination in this context. $\endgroup$
    – Arnaud Mortier
    May 13, 2018 at 20:30
  • $\begingroup$ Guassian elimination should not leave you only with the solution 0,0. If you post your elimination work/thoughts someone might be able to explain what you were doing/thinking wrong. It seems like your issue may be with Gaussian elimination on underdetermined systems, which is broader than finding eigenvectors. $\endgroup$
    – Mark S.
    May 14, 2018 at 14:02

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If you know what the eigenvalues are, for instance by finding the roots of the characteristic polynomial, then you can apply Gaussian elimination to find solutions to the system $$(M-\lambda \operatorname{Id})X=0$$ where $\lambda$ is one of your eigenvalues and $X$ is an unknown eigenvector associated with it.

Given that your question is more precisely when not to apply Gaussian elimination, as far as I can see there is no other situation than this one where Gaussian elimination would be useful.

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  • $\begingroup$ I suspect the OP made a mistake in applying Gaussian elimination to the system $Ax = 0$ mentioned in the Question, or did not appreciate the outcome in providing nontrivial solutions to that homogeneous problem. So it's a bit hard to know what advice is needed, with no evidence of what mistake or what misunderstanding occurred. $\endgroup$
    – hardmath
    May 13, 2018 at 21:07
  • $\begingroup$ Thank you. I have added a bit more details to clarify my confusion. If you could please give it a look. $\endgroup$
    – momo
    May 14, 2018 at 13:55

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