Recipe

Find the smallest k for a generalized eigenvector

Use when: you have a vector $\mathbf v$ and an eigenvalue $\lambda$ , and need to know whether $\mathbf v$ is a generalized eigenvector, or the smallest $k$ with $(T-\lambda I)^k\mathbf v=\mathbf 0$

Goal: the smallest exponent $k$ , plus the ordinary eigenvector that shows up one application before it

Taught in Generalized eigenvectors

Quick recipe

Apply $(T-\lambda I)$ once and check the result
Keep applying $(T-\lambda I)$ to the running result, one application at a time
Report the smallest $k$ , and note the intermediate result is an ordinary eigenvector

1
Compute $(T-\lambda I)\mathbf v$ ; if it's already $\mathbf 0$ , $\mathbf v$ is generalized with $k=1$ (an ordinary eigenvector); otherwise keep the result and continue.

Example

For $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ and $\mathbf w=(0,1)$ : $(A-2I)\mathbf w=(1,0)\ne\mathbf0$ , so continue to Step 2.
- $(T-\lambda I)\mathbf v=\mathbf0$ — stop here; $\mathbf v$ is generalized with $k=1$ .
- $(T-\lambda I)\mathbf v\ne\mathbf0$ — keep the nonzero result and continue.
Why: the definition itself is exactly this test at $k=1$ , so checking it first catches the easy ordinary-eigenvector case immediately.
2
Apply $(T-\lambda I)$ again to the PREVIOUS result, not to the original $\mathbf v$ recomputed from scratch, and check whether the new result is zero.

Example

Continuing with $(A-2I)\mathbf w=(1,0)$ : apply $(A-2I)$ again, $(A-2I)(1,0)=(0,0)$ , zero after $2$ total applications.
- the new result is $\mathbf0$ — the total number of applications used so far is the smallest $k$ ; stop.
- the new result is still nonzero — repeat this same step once more.
⚡ Shortcut

For a single $n\times n$ Jordan block with eigenvalue $\lambda$ (the superdiagonal-of- $1$ s form), the standard basis vectors $\mathbf e_1,\dots,\mathbf e_n$ have smallest $k=1,2,\dots,n$ in that same order, read off with no computation at all.

Why: $(T-\lambda I)^k\mathbf v$ is literally $(T-\lambda I)$ applied $k$ times in a row, so counting applications one at a time is exactly what the exponent means.
3
The count from Step 2 is the smallest exponent; the result one application before it hit zero satisfies the ordinary eigenvector equation.

Example

For $\mathbf w=(0,1)$ : $k=2$ , and the intermediate result $(A-2I)\mathbf w=(1,0)$ is itself an ordinary eigenvector of $A$ for $\lambda=2$ .

Why: the vector one step before the chain reaches zero satisfies $(T-\lambda I)\mathbf u=\mathbf0$ exactly, the ordinary-eigenvector condition itself.

Common mistakes

Recomputing $(T-\lambda I)^k\mathbf v$ from scratch at every exponent, instead of applying $(T-\lambda I)$ to the previous result.
Reporting a $k$ larger than necessary, without checking whether an earlier exponent already gave zero.
Assuming $(T-\lambda I)\mathbf v=\mathbf0$ for one vector says anything at all about a different, unrelated vector.
Treating "generalized" and "not ordinary" as synonyms, when every ordinary eigenvector is also generalized, with $k=1$ .