Given a Hermitian operator $L$ and 2 distinct eigenvectors $\lambda_1$ and $\lambda_2$ with the same eigenvalue $\lambda$ (confusing to use $\lambda$ for both but that's what the book does), why is the following statement obvious, or even true?
It should be clear that any linear combination of the two eigenvectors is also an eigenvector with the same eigenvalue.
It's clear to me that the linear combination $\lambda_1+\lambda_2$ would have the same eigenvalue of $\lambda$ because they must be orthogonal. But for example wouldn't the linear combination $2\lambda_1+3\lambda_2$ have a different eigenvalue?
I'm kind of learning linear algebra on the fly so I may be misreading the statement altogether. I'm trying to break things down into components but still not seeing how the above statement is true.
Let $A$ be an linear mapping and $x_1,x_2,\cdots, x_n$ be its eigenvectors with the same eigenvalue $\lambda$. Let $c_1,c_2,\cdots,c_n$ be scalars and put $x=\Sigma c_ix_i$. Then $Ax=\Sigma Ac_ix_i=\Sigma c_iAx_i=\Sigma c_i\lambda x_i=\lambda\Sigma c_ix_i=\lambda x$.