Can anybody tell me what this notation means?
$A|_{ker A}=0$
or this:
$A|_{im A}=A|_{ker(I-A)}=I$
Generally, I basically don't know what this notation means generally? Apparently a linear operator with such properties is diagonisable, but first I have to understand what this notation means. The notes just mention it with no explanation. Thanks
$\endgroup$2 Answers
$\begingroup$If $f$ is a function between sets $X \to Y$ and $Z \subseteq X$, then the restriction $f|Z$ is the function $Z \to Y$ given by $(f|Z)(z) = f(z)$.
Informally, a restriction keeps the rule but reduces the domain.
Note that this can have consequences. For instance, the restriction of $x \mapsto x^2$ to $(0,\infty)$ is injective but the original function with domain $\mathbb R$ is not.
In your linear algebra example, $A|(\ker A)=0$ means that $A$ sends every vector in $\ker A$ to the zero vector, which follows from the definition of $\ker A$. In other words, $A$ is the zero transformation in $\ker A$.
$\endgroup$ 4 $\begingroup$Let $M$ and $N$ non-empty sets and $f:M \to N$ a map.
If $A$ is a subset of $M$, then we can define the restriction$f_{|A} \to N$ of $f$ to $A$ by
$f_{|A}(a):=f(a)$ for all $a \in A$.
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