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I've started studied eigenvalues and eigenvectors.

If there is a transformation T: V->V

I can find out a matrix of T with fixed basis and characteristic polynomial of T.

With this characteristic polynomial of T, I can find out eigenvalues and eigenvectors

Does this mean that this transformation T: V->V has eigenvalues and eigenvectors?

And,

If there is a transformation T: S->V, S is a subspace of V.

then Matrix of T is not square, and therefore I can not find out characteristic polynomial. Does this imply that there exists no eigenvalues and eigenvectors under Transformation from subspace into Space ?

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To answer your first question: yes. If you can find the characteristic polynomial of a transformation, then its eigenvectors are the zeros of that polynomial. Note that these eigenvalues will be the same, regardless of your choice of basis.

As for your second question: eigenvalues (like the determinant) are only defined on transformations that take a space to itself (or to a space of the same dimension). So, $T:S \to S$ has eigenvalues, as does $T:V \to V$, but not $T:S \to V$ (even if $S$ is a subspace of $V$).

If $S$ is a $T$-invariant subspace of $V$, then we could take the restriction of the map $T: V \to V$ to get the "smaller" transformation $T:S \to S$. Also, if $S \subset V$ and $T:V \to S$, then we can think of $T$ as a (non-surjective) map from $V$ to $V$ and thereby find its eigenvalues.

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  • $\begingroup$ I appreciate your answer. One more thing about my first question, then for any transformation and any matrix with chosen basis with respect to that transformation, eigenvalues and eigenvectors are the same? $\endgroup$ – dad Oct 31 '14 at 2:08
  • $\begingroup$ And, for given transformation from V to V, if we want to find out eigenvalues and eigenvectors, finding out roots of characteristic polynomial is the only way to get those values? $\endgroup$ – dad Oct 31 '14 at 2:12
  • $\begingroup$ The eigenvectors depend on the basis. Finding the roots of the characteristic polynomial is not the only way, but it is the usual way. $\endgroup$ – Omnomnomnom Oct 31 '14 at 2:56
  • $\begingroup$ Forgot to include: eigenvalues do not depend on the basis $\endgroup$ – Omnomnomnom Oct 31 '14 at 3:03

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