Let $A$ be an $n \times n$ matrix, and $u, v$ be eigenvectors corresponding to an eigenvalue $\lambda$ of $ A$ (that is, $Au = \lambda u$ and $Av = \lambda v$). Is it true that $u + v$ is an eigenvector corresponding to the eigenvalue $\lambda$.

I know that an eigenvector can't be the $0$ vector, but an eigenvalue can be $0$, it just means the matrix $A$ is not invertible.

By definition $ Au = \lambda u$, where $u \ne 0$, and $Av = \lambda v$, where $v \ne 0$. So, $$Au + Av = \lambda u + \lambda v \implies A(u + v) = \lambda (u + v)$$ where $u + v \ne 0$?

I feel like I'm missing something because my solution was too straightforward.

  • $\begingroup$ It is that straight forward as the set of eigenvectors of the same eigenvalue (together with the zero vector) form a subspace called the eigenspace for that eigenvalue where every non zero vector has the same eigenvalue. $\endgroup$ – Random Excess Jul 28 '14 at 17:17
  • 2
    $\begingroup$ $u$ can be $-v$, but in any other case it's true $\endgroup$ – DGRasines Jul 28 '14 at 17:17
  • $\begingroup$ I reformatted the post. See math notation guide. $\endgroup$ – user147263 Jul 28 '14 at 19:38

Well, the set of all eigenvectors of an eigenvalue forms a subspace.

I.e. if u,v $\in Eig(A,\lambda)$ then $a_1 u$+$a_2v\in Eig(A,\lambda)$:

For $Av=\lambda v$ and $Au=\lambda u$ we have:

$A(a_1u+a_2v)=a_1Au+a_2Av=a_1\lambda u+a_2\lambda v=\lambda(a_1u+a_2v)$

Note: $0$ is not an eigenvector. So the set above is only a space if we add $0$ to the space.

  • $\begingroup$ The set of such eigenvectors only forms a subspace if you allow the zero-vector as an eigenvector. $\endgroup$ – Omnomnomnom Jul 28 '14 at 17:22
  • $\begingroup$ true. I will add it above $\endgroup$ – Marm Jul 28 '14 at 17:23

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