Examples of eigenspaces which are infinite dimensional
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Let V be the vector space of all real-valued sequences. Its dimension is countably infinite, i.e., it has the same cardinality as the integers.
Take the linear transformation T from V to V that just shifts everything to the left: it takes
(x_0, x_1, x_2, x_3, ...) to (x_1, x_2, x_3, x_4, ...)
If you try and work out the eigenvalues and eigenvectors of T, the definition of eigenvectors tells you that, given an eigenvalue c, a sequence is an eigensequence for c if and only if the terms of the sequence satisfy x_i = c x_{i-1}. Those are geometric sequences: they look like (1, c, c^2, c^3, ...).
What that means is that every real number is an eigenvalue for T, and has a 1-dimensional eigenspace. There are uncountably many eigenvalues, but T transforms a countably infinite-dimensional space to itself. In finite dimensions, you can't have more eigenvalues than the dimension of the space, and in infinitely many dimensions, the same doesn't even hold even if you use transfinite cardinals.
Take the linear transformation T from V to V that just shifts everything to the left: it takes
(x_0, x_1, x_2, x_3, ...) to (x_1, x_2, x_3, x_4, ...)
If you try and work out the eigenvalues and eigenvectors of T, the definition of eigenvectors tells you that, given an eigenvalue c, a sequence is an eigensequence for c if and only if the terms of the sequence satisfy x_i = c x_{i-1}. Those are geometric sequences: they look like (1, c, c^2, c^3, ...).
What that means is that every real number is an eigenvalue for T, and has a 1-dimensional eigenspace. There are uncountably many eigenvalues, but T transforms a countably infinite-dimensional space to itself. In finite dimensions, you can't have more eigenvalues than the dimension of the space, and in infinitely many dimensions, the same doesn't even hold even if you use transfinite cardinals.
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