# countable dimensional vector space has uncountable eigenvalues?

Consider $$\mathcal{R}^{\infty}$$, and linear map $$\mathcal{L} \in L(\mathcal{R}^{\infty})$$, where $$\mathcal{L}((x_1,x_2,...))=(x_2,x_3,...)$$. Now, any number $$\lambda \in \mathcal{R}$$ is an eigenvalue with eigenvector $$(c,c\lambda,c \lambda^2,...)$$. But dimention of $$\mathcal{R}^{\infty}$$ is countable. Is this a problem?

• @AndrewOstergaard: Not mine the downvote, but I suppose that it is motivated by some confusion in your answer between a Hamel basis ( that in this case is not countable) and a Schauder basis as your $e_1,e_2,...,e_n,...$. See: math.stackexchange.com/questions/630142/… – Emilio Novati Feb 1 '20 at 21:57

A basis of a vector space (called a Hamel basis) is a set of linearly independent vectors such that any vector can be expressed as a linear combination of elements of the basis. But note that a linear combination is a finite sum of products of scalars and vectors. So, in your case the set of vectors $$E=\{(1,0,0,\cdots),(0,1,0,0\cdots),(0,0,1,0,0\cdots)\}$$ is not a basis, because we cannot express all sequences as a finite linear combination of those sequences.
So it is wrong that the dimension of $$\mathbb{R}^\infty$$ is conuntable. And we can have an uncountable set of linearly independent eigenvectors.
If we think at $$\mathbb{R}^\infty$$ as a normed vector space than the set $$E$$ is a Schauder basis (and any vector can be expressed as an ''infinite linear combination''of elements of $$E$$) , but the vector dimension of the space remains uncountable infinite.