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Suppose that there is a linear map $L$ such that:

$$ L:(U,F)\to (U,F) $$

Where $(U,F)$ is a linear space. Now, matrices are merely representations of a linear map with respect to a given basis - for example matrix $A\in F^{\dim U \times \dim U}$ could be the matrix representation of $L$ with respect to the canonical basis. We know how to find eigenvalues of the matrix $A$.

However, is there such a thing as eigenvalues of the actual linear map $L$ and not of its matrix representation? If so, how do you find them?

Specifically: we can find eigenvalues of a matrix by finding the zeros of the characteristic polynomial. Is it possible to find the characteristic polynomial of a linear map without needing to first define a basis with respect to which we can write a matrix representation? If not, is it possible to find the eigenvalues with such an algorithmic method, without having to define a basis first?

Thanks for your insight!

space_voyager
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    If $(U,F)$ is finite dimensional, then the situation is the same as it is for matrices (though the field not being algebraically closed can come back to bite you). If $(U,F)$ is infinite dimensional then the situation is much more subtle. – Ian Oct 10 '15 at 21:53
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    Actually eigenvectors are defined for linear maps first. Matrices are simply means to represent these linear maps w.r.t. a given basis, and to make computations about them.. – Bernard Oct 10 '15 at 21:54
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    An eigenvalue of $L$ is just a non-zero element $u\in U$ for which there exists a non-zero $\lambda\in F$ such that $L(u)=\lambda u$. – Zev Chonoles Oct 10 '15 at 21:55
  • I added the Specifically section to perhaps get a more direct answer to my question. Thanks for your thoughts so far! – space_voyager Oct 10 '15 at 21:59
  • Can you give an example? The problem is I'm having a hard time conceptualizing what this means. I was always used to working with matrices, so I cannot see how to represent $L$ other than with a matrix... – space_voyager Oct 10 '15 at 22:21
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    I can't think of any way of characterizing a linear transformation without knowing how it transforms a basis. If you know a way, come up with an example and I'll see if I can figure out how to find the characteristic polynomial for it. Of course, if you're given a basis and the image of those basis vectors under your transformation, then any method you choose to find the eigenvalues (and there are other methods) will be equivalent to the matrix approach. –  Oct 10 '15 at 22:29
  • I took a "math for engineering" class and this sort of thing was done; lots of eigenvalues and eigenfunctions with nary a matrix in sight. Example 1 in this link is the sort of thing I have in mind. – pjs36 Oct 10 '15 at 22:31
  • @pjs36 if you can give an example in an answer, that'd be great. My eyes hurt reading Paul's website and there's too much on that page for the "simple" "mind-settling" explanation that I'm searching for. – space_voyager Oct 10 '15 at 22:33

2 Answers2

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Not sure this is exactly what you want, but it's a linear transformation without explicitly specifying a matrix, or action on a basis, and also finding (real) eigenvalues.

Given a vector $v$ in ${\bf R}^3$, define the linear transformation $T_v:{\bf R}^3\to{\bf R}^3$ by $T_v(w)=v\times w$ (the cross product).

Now suppose $T_v(w)=\lambda w$. We know $T_v(w)$ is orthogonal to $w$, so we have to have $\lambda w$ orthogonal to $w$, which (for $w\ne0$) can only happen if $\lambda=0$. So, we have found the only real eigenvalue, zero (and a corresponding eigenvector is $v$).

Gerry Myerson
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  • Thanks, I guess this is a simple application of what I'm asking. I was really asking for a general method, as what you did will not apply to a different transformation. But thanks anyhow! – space_voyager Oct 10 '15 at 22:49
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    I think a general method is too much too ask for, as there are so many different ways in which a linear transformation can be presented. – Gerry Myerson Oct 10 '15 at 22:54
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Another example: consider the linear transformation on the vector space of infinitely differentiable functions given by $T(f)=f'$. You get the eigenvalues (and eigenvectors) by solving the differential equation $${dy\over dx}=\lambda y$$ Every real number $\lambda$ is an eigenvalue, with corresponding eigenvector $e^{\lambda x}$.

Gerry Myerson
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