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Learning Differential topology, Sorry for asking anything trivial. I am stuck in this question:

Let $V$ be a vector space and let $\Delta$ be the diagonal of $V \times V$ . For a linear map $A : V \to V$ consider the graph $W = \{(v, Av) \in V \times V \}$. Show that $W \pitchfork \Delta$($W$ transversal to $\Delta$) if and only if $+1$ is not an eigenvalue of $A$.

My attempt:

So, $\Delta \cap W=\{(v,v) \mid Av=v\}.$ If +1 is an eigenvalue of $A$, then $\det(A-I)=0$ which implies $A-I$ has rank less than $n$. So, for $A-I:V \to V$, the derivative at any point $p\in V$ is zero that is, $D(A-I)(p)=0$ which implies $D(A-I)$ is not surjective at any point $p\in V$. So, $p$ is a critical point of $A-I$. but then how can I use it to conclude that $\Delta$ is not transverse to $W$?

For the other way, suppose $+1$ is not an eigenvalue of $A$, then $\Delta\cap W=\{(0,0)\}$, as $Av=v\implies v=0$(because 1 isn't an eigenvalue). So, $\Delta$ and $W$ intersects only at $(0,0)$. And hence they are transversal being no intersecting tangentially. Is that a correct argument?

Brian61354270
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Infinity
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1 Answers1

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With the work you have done, you can prove both implications simultaneously. You already showed that $$ \Delta\cap W=\{(v,v):Av=v\}=\{(v,v):v\in\text{Ker}(A-I)\}. $$ So we get \begin{align} W\pitchfork\Delta & \Leftrightarrow W+\Delta=V\times V\\ &\Leftrightarrow W\cap\Delta=\{(0,0)\} \\ &\Leftrightarrow \text{Ker}(A-I)=\{0\}\\ &\Leftrightarrow A-I \ \text{is invertible}\\ &\Leftrightarrow \det(A-I)\neq 0\\ &\Leftrightarrow 1\ \text{is no eigenvalue of}\ A. \end{align}

studiosus
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