$A$ is a $4\times 4$ matrix that has an eigenvalue $\lambda.$ Prove that if $v = (18, 6, 7, 100)$ and $w = (5, -1/2, -2, 1)$ are eigenvectors associated with the eigenvalue $\lambda,$ so is the vector $2v-w.$

All I've gotten so far is that $2v-w = (31, 23/2, 16, 199)$ and I was thinking if I could use the property of $A(cv) = c(Av) = c(\lambda v) = \lambda(cv)$ to find the eigenvalue somehow but I got stuck and I'm not sure if it is necessary when solving this.

  • $\begingroup$ If you want to improve your TeX a bit, please have a look at the code for your question now (you can see it by clicking "edit"). In general, you should include the entire math expression with dollar signs, not just the letters in your expression. Cheers. $\endgroup$ Oct 13, 2017 at 11:28
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    $\begingroup$ \begin{align} Av&=\lambda v.\\ Aw&=\lambda w .\\ A(2v-w)&=2Av-Aw \\ &=2\lambda v-\lambda w \\ &= \lambda(2v-w). \end{align} (But see @MarcvanLeeuwen's answer for the details.) $\endgroup$ Oct 13, 2017 at 12:35

1 Answer 1


It is actually false that $2v-w$ is always an eigenvector for$~\lambda$, for a subtle reason of definition. But in the concrete case given, it is an eigenvector, because it is nonzero. The set of eigenvectors for$~\lambda$ together with the zero vector which is not an eigenvector form the eigenspace $\ker(A-\lambda I)$ for$~\lambda$, which as the name suggests, and as any kernel of a linear map, is a subspace of your vector space. Linear (sub)spaces are closed under forming linear combinations.

So $2v-w$ is definitely an element of the eigenspace for$~\lambda$. However, it is not guaranteed to be nonzero. Indeed, taking $v$ to be any eigenvector for$~\lambda$ and setting $w=2v$ gives an example where $2v−w=0$ is not an eigenvector. However for the concrete given values for $v$ and $w$, this does not happen.

The reason why the zero vector is not considered to be an eigenvector, is that one wants to be able to associate an eigenvalue to each eigenvector; if allowed, the zero vector would be eigenvector for any $\lambda$ whatsoever, all at once.

  • $\begingroup$ Much better answer than mine, which I made into a comment instead. $\endgroup$ Oct 13, 2017 at 12:35

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