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Two little questions to this passage:

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(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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Salamo Mar 23 at 7:47
  • $\begingroup$ 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 $f \circ T = 0$. $\endgroup$ – Joshua Mundinger Mar 23 at 14:42
  • $\begingroup$ There was also a typo in the definition of $T$, which is now fixed. $\endgroup$ – Joshua Mundinger Mar 23 at 14:52
  • $\begingroup$ 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? $\endgroup$ – Salamo Mar 23 at 19:26
  • $\begingroup$ 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?) $\endgroup$ – Joshua Mundinger Mar 24 at 1:25

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