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If we have a 2D plane and the hermitian matrix $L$ where: $$L|\lambda_1\rangle=\lambda|\lambda_1\rangle$$ $$L|\lambda_2\rangle=\lambda|\lambda_2\rangle$$ Given that $|\lambda_1\rangle$ and $|\lambda_2\rangle$ are linearly independent, we can make any vector $$|A\rangle=\alpha|\lambda_1\rangle +\beta|\lambda_2\rangle$$ that will be an eigenvector for $L$ with the same eigenvalue. Can't I, with the last equation, create every vector in that plane and therefore every vector is an eigenvector for the matrix with the same eigenvalue? I know this question might sound ridiculous and that there is a mistake in my reasoning but I can't find where is the mistake.

zizaaooo
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There is no mistake. If you have a $2 \times 2$ matrix with one eigenvalue $\lambda$ and 2 linearly independent eigenvectors, then the whole plane is the eigenspace and the matrix is equal to $\lambda I$.

Vladimir Lysikov
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  • So if we are talking about an ordinary 2D plane the hermitian operator only stretches that plane without flipping or shearing it? – zizaaooo Dec 29 '23 at 18:56
  • A hermitian operator can have two different eigenvalues, for example, $\begin{bmatrix}1 & 0 \ 0 & 2\end{bmatrix}$ has two eigenvalues $\lambda_1 = 1$, $\lambda_2 = 2$. In this case not every vector is an eigenvector. If you only have one eigenvalue, this is a very special situation. – Vladimir Lysikov Dec 29 '23 at 18:57
  • But isn't this special situation the one presented in my question? And in that situation doesn't the matrix stretches the plane evenly in every direction? – zizaaooo Dec 29 '23 at 19:58
  • In your question you write $L\left|\lambda_1\right> = \lambda \left|\lambda_1\right>$, $L\left|\lambda_2\right> = \lambda \left|\lambda_2\right>$, and I read it as two linearly independent vectors with the same eigenvalue $\lambda$.

    Is that not correct? Should that have been $L\left|\lambda_1\right> = \lambda_1 \left|\lambda_1\right>$, $L\left|\lambda_2\right> = \lambda_2 \left|\lambda_2\right>$?

     – Vladimir Lysikov Dec 29 '23 at 20:00
  • Yess that is correct. It not $\lambda_1, \lambda_2$ it is only $\lambda$ – zizaaooo Dec 29 '23 at 20:01
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    If the eigenvalue is the same, then everything you write is correct. But not every Hermitian operator will have only one eigenvalue. – Vladimir Lysikov Dec 29 '23 at 20:03