Intuition of generalized eigenvector.

Question

I was trying to get an intuitive grasp about what the the generalized eigenvector intuitively is. I read this nice answer, so I understand that in the basis given by the generalized eigenvectors, a jordan block is a linear map that is the sum of a stretch by a factor $\lambda$ (eigenvalue associated to the block) and a "collapse", but I don't understand the conclusion on what these famous generalized eigenvectors actually are...

Thus the kernel of $(T−λI)k$ picks up all the Jordan blocks associated with eigenvalue $λ$ and, speaking somewhat loosely, each generalized eigenvector gets rescaled by $λ$, up to some "error" term generated by certain of the other generalized eigenvectors.

Maybe someone that actually understand the last argument can care to explain with some more detail? Thank you.

Answer 1

Don't look for anything particularly deep or fancy here.

If you have a calculation to do that involves some vectors and a linear operator $T$ that you apply perhaps to several of the vectors or more several times in sequence, then it can simplify the calculation if you represent the vectors in an eigenbasis -- since then we have $$ T(x_1,x_2,\ldots,x_n) = (\lambda_1x_1, \lambda_2x_2, \ldots, \lambda_n x_n) $$ Each component of the vector just gets multiplied by its associated eigenvalue and the different components don't interact with each other at all.

Unfortunately, not all operators can be written in this nice form, because there may not be enough eigenvectors to combine into a basis. In that case the "next best thing" we can do is choosing a basis where the matrix of $T$ is in Jordan form. Then each component of $T(x_1,x_2,\ldots,x_n)$ is either $\lambda_i x_i$ or $\lambda_i x_i + x_{i+1}$, which is still somewhat simpler than multiplication by an arbitrary matrix.

Since this gives us some of what a basis consisting entirely of eigenvectors gives us, in terms of computational simplicity, if seems reasonable to describe them as a generalization of eigenvectors. Especially since in the case where we do have enough eigenvectors for an eigenbasis, generalized eigenvectors are the same as eigenvectors.

Answer 2

I see generalized eigenvectors as an attempt to patch up the discrepancy between geometric multipicity and algebraic multiplicity of eigenvalues. This discrepancy is most easily observed as a result of shear transformations (and looking at Jordan normal forms, we see that shear transformations are in some sense at the core of any such discrepancy).

For instance, take the shear transformation given by the matrix $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} $$ It has eigenvalue $1$ with algebraic multiplicity $2$ (the characteristic polynomial is $(\lambda - 1)^2$, which has a double root at $1$), but geometric multiplicity $1$ (the eigenspace has dimension $1$, as it just the $x$-axis).

However, the space of generalized eigenvectors is the entire plane, which is 2-dimensional, and more in line with the algebraic multiplicity of the eigenvalue.

Answer 3

Consider the matrix $$A= \begin{bmatrix}3 & 1 \\ 0 & 3 \end{bmatrix}$$. Obviously 3 is the only eigenvalue with "algebraic multiplicity" 2. It is also easy to find the eigenvectors $$\begin{bmatrix}3 & 1 \\ 0 & 3 \end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}= \begin{bmatrix}3x+ y \\ 3y\end{bmatrix}= \begin{bmatrix}3x \\ 3y \end{bmatrix}$$ so that we have 3x+ y= 3x and 3y= 3y. The first equation gives y= 0 and the second is satisfied for any x. So any eigenvector is of the form $$\begin{bmatrix}x \\ 0 \end{bmatrix}= x\begin{bmatrix}1 \\ 0 \end{bmatrix}$$. So the subspace of all eigenvectors has dimension 1 (the geometric multiplicity is 1).

$$v= \begin{bmatrix}0 \\ 1 \end{bmatrix}$$ is NOT an eigenvector: $$(A- 3I)v= \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix}= \begin{bmatrix}1 \\ 0\end{bmatrix}$$, NOT the zero vector. But it does give the previous eigenvector so that applying A- 3I again would give the 0 vector it is a "generalized eigenvector.

Answer 4

First, I want to introduce the Jordan–Chevalley decomposition:
In essence it formalizes what you said: For every trigonalizable map $f \in End(V)$ with transformation Matrix $A$, there exists a diagonal Matrix $D$ (the stretching part) and an nilpotent Matrix $N$ (the “collapse” part) such that, $A = D + N$
(which makes intuitive sense: $D$ will have the same Main diagonal as $A$, so $A – D$ will be an strictly triangular matrix $\Leftrightarrow$ nilpotent).

Now consider a Matrix $B$ with just one Eigenvalue $\lambda$:
Because $\lambda$ is the sole Eigenvalue of $B$, the diagonal Matrix $D$ in the Jordan–Chevalley decomposition will just have $\lambda$ on its diagonal, so: $\lambda * \mathbb{I}_n = D$.
Now we rearrange:

$B = D + N$ to $B - D = N \Leftrightarrow (B - \lambda * \mathbb{I}_n) = N$.

Because $N$ is nilpotent, it follows that for an sufficient $k$, every Vector in the generalized Eigenspace will be in the Kernel of $(B - \lambda * \mathbb{I}_n)^k$, because for a sufficient $k$, $(B - \lambda * \mathbb{I}_n)^k$ will be the zero map.

Important: If you have more than one Eigenvalue, you need to restrict the Matrix to the respective generalized Eigenspace to reduce it to the above situation.
In this case, $(A - \lambda * \mathbb{I}_n)^k$ will not map every vector in V to zero, but just Elements of the generalized Eigenspace associated to $\lambda$ (Because you applied the Jordan–Chevalley decomposition to the generalized Eigenspace and not to V)
This is what he meant by “$(A - \lambda * \mathbb{I}_n)^k$ picks up all the Jordan blocks associated with eigenvalue Lambda“.

I think the "error" term he wrote about, are the Jordanchains / the way you calculate them:
$A * v_j = v_{j-1} + \lambda * v_j \Leftrightarrow v_j * (A - \lambda * \mathbb{I}_n) = v_{j-1}$
So every Vector not only gets scaled, but you also add a Vector. This added Vector is the “error Term".