# How doesn't combining two eigenvectors that have the same eigenvalue for a specific matrix represent every vector left in the plane?

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.

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 $$\lambda I$$.
• 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. Commented Dec 29, 2023 at 18:57
• 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 $L\left|\lambda_1\right> = \lambda_1 \left|\lambda_1\right>$, $L\left|\lambda_2\right> = \lambda_2 \left|\lambda_2\right>$? Commented Dec 29, 2023 at 20:00