# Linear combination of eigenvectors corresponding to the same eigenvalue

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?

• You are already half there. If u is a linear combination of eigenvectors, what happens when you multiply by A? – Matthew Leingang Dec 11 '14 at 22:11
• Ok I seem to have understood from your answer and the answer below. The following part states 'Conclude that any non-zero vector u which is an element of the span of v1,...vk is an eigenvector of A corresponding to the eigenvalue Lamda'. Doesn't this just automatically follow? If U is a linear combination of the eigenvectors v1,...vk then it must be within the span of v1,...,vk and by definition an eigenvector is if Au=Lamdau hence it must be an eigenvector of A with eigenvalue Lamda? – Orangegulf Dec 11 '14 at 22:33
• @Orangegulf Yes, "conclude that" could be replaced by "you have just proved that". – augurar Dec 11 '14 at 22:51

For each $v_i$, $Av_i = \lambda v_i$. Suppose $u = \sum_i{c_i v_i}$. Since matrix multiplication is linear, $$Au = A \sum_i {c_i v_i} = \sum_i{c_i Av_i} = \sum_i {c_i \lambda v_i} = \lambda \sum_i {c_i v_i} = \lambda u$$