Vectors that are not the eigenvectors of any linear operator

Question

Let me preface this by saying that I am not a mathematician, so forgive me if my phrasing is not rigorous or not formally correct.

An eigenvector is some vector that when acted on by a linear operator produces that some vector multiplied by some scalar constant (the eigenvalue). So evidently, a defined linear operator has a finite a number defined eigenvectors and values. All other vectors are not eigenvectors of THIS linear operator. However, is it true to say that these vectors are necessarily eigenvectors of SOME OTHER defined linear operator? In other words; is it true to say that all vectors are eigenvectors of some linear operator (evidently not the same one), or are some vectors eigenvectors of no linear operator?

Thanks

Answer 1

Yes every non sero vector is an eigen vector for some matrix.

Note that the matrix $\lambda I$ has every nonzero vector as its eigenvector with the eigenvalue $\lambda$

Answer 2

Every vector is an eigenvector for the identity transformation with eigenvalue $\lambda =1$, as $I\mathbf{v}=\mathbf{v}=1\mathbf{v}$. I suppose the single exception is the zero vector, which is excluded for technical reasons.

Answer 3

If you are in an inner-product space (i.e., a vector space that has an inner product as well, such as $\Bbb R^n$), and $u$ is any nonzero vector, then the transformation $$ v \mapsto v - \frac{v \cdot u}{u \cdot u} u $$ is a non-identity linear transformation for which $u$ is an eigenvector (with associated eigenvalue zero).

NB: Any finite-dimensional inner product space $V$ can be endowed with an inner product via an isomorphism to $\Bbb R^n$, where $n = dim V$, so this is a pretty general construction.

Answer 4

Start with any nonzero vector $v$ you like. Find $n-1$ more vectors that together with $v$ form a basis. Then choose any eigenvalue $\lambda$ you like, define $Tv = \lambda v$ and define $T$ any way you like on the rest of the basis.

This is, in a sense, the most general solution. You set $Tv = \lambda v$ and then define $T$ arbitrarily on a subspace complementary to the line spanned by $v$.

Even without this explicit construction, the answer to your question would have to be "no" - every vector is an eigenvector for some linear transformation. That's because every nonzero vector in an abstract vector space is just the same as any other. Once one of them has some property, so do they all.