# Normalizing dual eigenvectors? Why only for trivial defects?

Two little questions to this passage:

(1) How can we normalize to get $$\langle u,u^*\rangle =1$$?

(2) Why is this possible if $$\lambda$$ has trivial defect only (i.e. for trivial Jordan blocks)? I did not get it from the text.

## 1 Answer

The normalization is not deep: just that if $$\langle u, u^\ast\rangle \neq 0$$, then $$u^\ast$$ may be rescaled so that that inner product is 1. Asking whether we can find $$u^\ast$$ such that $$\langle u, u^\ast \rangle \neq 0$$ is deeper.

I'm not sure whether you're working with a complex vector space with a Hermitian inner product, or more generally with a linear transformation and its dual. Below, I'll give an exposition of the latter; the former can be obtained by replacing duality with orthogonality, along with an appropriate sprinkling of complex conjugates.

Let $$T: V \to V$$ be an endomorphism of a finite-dimensional vector space $$V$$ over an algebraically closed field $$k$$, and let $$T^*: V^* \to V^*$$ denote the dual transformation. Let the Jordan blocks of $$T$$ acting on $$V$$ be $$V = V_1 \oplus \ldots \oplus V_m$$. Then the action of $$T^*$$ has Jordan decomposition $$V^* = V_1^* \oplus \cdots \oplus V_m^\ast$$, with the same eigenvalues. Thus, to understand the relationship between the eigenvectors of $$T$$ and the eigenvectors of $$T^\ast$$, it suffices to analyze just one Jordan block. Further, the eigenvectors of $$T$$ and $$T - \lambda \, \mathrm{id}$$ are the same, so it suffices to consider when $$T$$ has eigenvalue $$0$$, i.e. is nilpotent.

In the case when $$T$$ is nilpotent and has one Jordan block, $$V$$ has a basis $$\{e_1,\ldots, e_n\}$$ with $$T e_i = e_{i-1}$$ for $$i > 1$$ and $$Te_1 = 0$$. That is, $$T$$ has matrix $$\begin{bmatrix} 0 & 1 &&& \\ & 0 & 1 & & \\ & & \ddots & \ddots & \\ & & & 0 & 1 \\ & & & & 0 \end{bmatrix}$$ Then $$T^\ast$$ has only a one-dimensional eigenspace, spanned by $$f$$ defined by $$f(e_i) = \begin{cases} 0 & i < n \\ 1 & i = n.\end{cases}$$ In terms of matrices, we can check this by taking the transpose of the matrix above. Then $$f$$ is our dual eigenvector and $$e_1$$ is our eigenvector, which has $$f(e_1) \neq 0$$ if and only if $$n = 1$$, i.e. $$T$$ has trivial defect.

• I see that $T^*$ has only one-dimensional eigenspace. But I don’t understand why it is spanned by the eigenfunction f which you have defined. – Salamo Mar 23 at 7:47
• The transpose has kernel spanned by $(0, \ldots, 0 ,1)$, which corresponds to the element in the dual basis of $\{e_1,\ldots, e_n\}$ corresponding to $e_n$, which is $f$. Alternatively, you can check that $f \circ T = 0$. – Joshua Mundinger Mar 23 at 14:42
• There was also a typo in the definition of $T$, which is now fixed. – Joshua Mundinger Mar 23 at 14:52
• Maybe one last question. Suppose $\langle u,u^*\rangle\neq 0$. You say then it’s no deep thing to scale $u^*$ such that $\langle u,u*\rangle=1$. Why and how? – Salamo Mar 23 at 19:26
• We have $\langle u, cv\rangle = c \langle u, v\rangle$ for all $c \in k$, and $v$ is an eigenvector if and only if $cv$ is (for $c \neq 0$), so if $v$ is an eigenvector we can set $u^\ast = cv$ such that $\langle u, u^\ast\rangle = 1$ (what value of $c$ would this have to be?) – Joshua Mundinger Mar 24 at 1:25