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?

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    $\begingroup$ @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/… $\endgroup$ – Emilio Novati Feb 1 '20 at 21:57

As noted in my comment, here, the main concept, is that of basis of a vector space and, as a consequence, the concept of dimension.( see the link in the comment)

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.


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