1. Is it true that a vector space V is a direct sum of all its eigenspace? What happens if T is not diagonalisable? Does this only apply to a vector space over an algebraically closed field?

  2. Similar to question 1, but now what if we are talking about generalised eigenspace? Will the answer be different?

  3. Is the multiplicity of the generalised eigenspace the same as the geometric multiplicity of the normal eigenspace?

  4. Can we always find a generalised eigenspace that does not only contain the zero vector?

  5. Is it true that we can always find a basis of a vector space V consisting of generalised eigenvectors of T?

  • 2
    $\begingroup$ Your question seems to be missing some linear transformations. $\endgroup$
    – Lee Mosher
    Oct 17, 2014 at 2:38
  • $\begingroup$ I am asking for a general case $\endgroup$ Oct 17, 2014 at 2:49
  • $\begingroup$ What does it mean "[vector space] $V$ is not diagonalisable"? I've heard of matrix/operator being diagonalisable, but vector space... Btw, Jordan decomposition answers some of these (generalized eigenvectors are columns of the similarity matrix $P$). As for 4, eigenvectors are never zero, by definition, so an eigenspace cannot contain only zero. I think you need to get more familiar with the terms you're using, because you seem a bit unsure of their meanings. $\endgroup$ Oct 17, 2014 at 3:29
  • $\begingroup$ sorry typo, i changed it to operator now $\endgroup$ Oct 17, 2014 at 4:05
  • $\begingroup$ i am indeed quite confused thats why i am asking $\endgroup$ Oct 17, 2014 at 4:06

1 Answer 1


For an arbitrary trasformation $T:V \to V$:

We say that $T$ is diagonalizable if and only if $V$ can be written as the direct sum of the eigenspaces of $T$, which is equivalent to finding a basis of $V$ consisting of eigenvectors. Not all transformations are diagonalizable.

If $V$ is a vector space over an algebraically closed field, then we can always write $V$ as the direct sum of the generalized eigenspaces of $T$. That is, over algebraically closed fields, we can always find a basis consisting of the generalized eigenvectors of $T$. If $V$ is defined over a field that is not algebraically closed, we have no such guarantee.

The multiplicity of the generalized eigenspace is equal to the algebraic multiplicity of the eigenvalue. This is equal to the geometric multiplicity of the eigenvalue (i.e. the dimension of the eigenspace) if and only if $T$ is diagonalizble.

We can always find an eigenvector for any linear transformation $T$ if our field is algebraically closed. We can also break $V$ up into the generalized eigenspaces of $T$ if our field is algebraically closed. If our field is not algebraically closed, we can't guarantee any non-trivial eigenspaces or generalized eigenspaces.

  • $\begingroup$ I am not sure if it is too much to ask for a proof here. But it will be good if you can provide one. Thanks anyway. $\endgroup$ Oct 18, 2014 at 2:13
  • $\begingroup$ I'm not sure what it is that you'd like me to prove $\endgroup$ Oct 18, 2014 at 3:30

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