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I followed this tutorial from this question. The dataset and example used in the mentioned tutorial seems famous one and it gives following eigen vectors:-

-.735178656   -.677873399
.677873399    -.735178656

I have used following formulas:-

enter image description here

i get following eigen vectors from both formulas:-

.735178656    .677873399
-.677873399   .735178659

Why is there difference in signs? Can any one explain. I have done all my research but could not find how come 3 x values in eigen vector be negative.

jaykio77
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    Welcome to the site! Think about the definition of an eigenvector for $\mathbf{A}$: $\mathbf{A}\mathbf{x} = \lambda\mathbf{x}$. What happens if we plug in $-\mathbf{x}$ instead? Does it still obey the "scaling rule"? – John Madden Nov 22 '22 at 18:30
  • I have used formula Ax=λx. No matter what, my answer never gets as that of i found on net (3 negative 1 positive). The duplicate question does not answer my question. – jaykio77 Nov 22 '22 at 18:41
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    Perhaps you could show us how you go from $\mathbf{A}\mathbf{x}=\lambda\mathbf{x}$ to an eigenvector? – John Madden Nov 22 '22 at 18:56
  • Whenever $u$ is a scalar for which $\mathbf A u=u\mathbf A$ and $u\lambda=\lambda u,$ then $$\mathbf A(u\mathbf x)=u\mathbf A x=u\lambda\mathbf x=\lambda(u\mathbf x)$$ demonstrates $u\mathbf x$ is an eigenvector with eigenvalue $\lambda.$ When you also require $|\mathbf x|^2=1,$ then $1=|u\mathbf x|^2=|u|^2|\mathbf x|^2=|u|^2.$ Among the many applications are that when the scalars are real numbers, $u=\pm1$ works, showing that whenever $\mathbf x$ is an eigenvector with eigenvalue $\lambda,$ then so is $-\mathbf x.$ (The implications for Complex and Quaternion scalars are more interesting.) – whuber Nov 22 '22 at 19:41

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