If v is an eigenvector of A with eigenvalue λ ("a λ -eigenvector") then so is kv for any k∈𝐂 .
34. Eigenspaces
34. Eigenspaces
In all the examples we've seen so far, the eigenvectors have all had a free variable in them. For example, in the last video, we found the eigenvectors for the matrix A=(32523-12-32-3112) to be:
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for λ=1 , (x-5x4x) ,
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for λ=2 , (x-x0) ,
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for λ=-1 , (x-xx)
For the matrix (2111) we found the eigenvalue 3±√52 had eigenvectors (x-1±√52x) . All of these have the free parameter x .
This is a general fact:
A(kv)=kAv=kλv=λ(kv) .
So for example, the vectors (x-xx) are all just rescalings of (1-11) . Indeed, people often say things like "the eigenvector is (1-11) ", when they mean "the eigenvectors are all the rescalings of (1-11) ". If you write this kind of thing in your answers, that's fine.
Suppose we have the matrix I=(1001) . The characteristic polynomial is det(I-λI)=det(1-λ001-λ)=(1-λ)2 , so λ=1 is the only eigenvalue. Any vector v satisfies Iv=v , so any vector (xy) is a 1 -eigenvector. This has two free parameters, so it is an eigenplane, not just an eigenline: there is a whole plane of eigenvectors for the same eigenvalue.
The set of eigenvectors with eigenvalue λ form a (complex) subspace of 𝐂n (i.e. closed under complex rescalings and under addition).
Let Vλ be the set of λ -eigenvectors of A . If v∈Vλ then kv∈Vλ (as we saw above). If v1,v2∈Vλ then A(v1+v2)=Av1+Av2=λv1+λv2=λ(v1+v2), so v1+v2 is also a λ -eigenvector.
We call this subspace the λ -eigenspace. In all the examples we saw earlier (except I ), the eigenspaces were 1-dimensional eigenlines (one free variable). So a matrix gives you a collection of preferred directions or subspaces (its eigenspaces), which tell you something about the matrix (e.g. if it's a rotation matrix, its axis will be one of these subspaces).
For the example A=(2111) we found the eigenvalues 3±√52 and eigenvectors (x-1±√52x) . We now draw these two eigenlines (in red).

Note that these eigenlines look orthogonal; indeed, you can check that they are! You do this by taking the dot product of the eigenvectors (it's zero). This is true more generally for symmetric matrices (i.e. matrices A such that A=AT ).