Suppose that $A$ is an element of $M_n(\mathbb{R})$ and $v_1,\dots,v_k\in\mathbb{R}^n$ are eigenvectors of $A$ corresponding to the same eigenvalue $\lambda$. Prove that if $u$ is a linear combination of the eigenvectors $v_1,\dots,v_k$, then $Au=\lambda u$.
I'm unsure of how to prove this. I understand that it is asking to show that some eigenvectors can have repeat eigenvalues, and that the linear combination of this therefore is also an eigenvector, but I can't seem to grasp the proof. Can someone give me a hint as to what to do?
Best Answer
For each $v_i$, $Av_i = \lambda v_i$. Suppose $u = \sum_i{c_i v_i}$. Since matrix multiplication is linear, $$Au = A \sum_i {c_i v_i} = \sum_i{c_i Av_i} = \sum_i {c_i \lambda v_i} = \lambda \sum_i {c_i v_i} = \lambda u$$