Exercise 173Linear algebra › Linear mapsL2Added 3 Nov 2026

Induced map on a quotient

#quotient space#induced map#well-defined

Statement

Let $f:E\to E$ be a linear endomorphism and $F$ an $f$ -invariant subspace. Recall that the quotient $E/F$ is the set of classes $[x]=x+F$ , with $[x]=[y]\iff x-y\in F$ , and vector space operations $[x]+\lambda[y]=[x+\lambda y]$ . Show that $f$ induces a well-defined linear map $\bar f:E/F\to E/F$ by $\bar f([x])=[f(x)]$ .

Hint

Show that if $[x]=[y]$ , then $[f(x)]=[f(y)]$ .

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Solution

Well-defined. Suppose $[x]=[y]$ . Then $x-y\in F$ . Since $F$ is $f$ -invariant, $f(x-y)=f(x)-f(y)\in F$ , i.e. $[f(x)]=[f(y)]$ . So the class $[f(x)]$ depends only on the class $[x]$ , and $\bar f$ is well-defined.

Linearity.
$\bar f\big([x]+\lambda[y]\big)=\bar f\big([x+\lambda y]\big)=[f(x+\lambda y)]=[f(x)+\lambda f(y)]=[f(x)]+\lambda[f(y)]=\bar f([x])+\lambda\bar f([y]).$

Conclusion. $\bar f$ is a well-defined linear endomorphism of $E/F$ , called the map induced by $f$ on the quotient. Invariance of $F$ is exactly what is needed for well-definedness.

Remark. In the block form of the previous exercise, $\bar f$ is represented by the block $A''$ .

Induced map on a quotient

Statement

Solution

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