# Linear map from quotient space to another space is well-defined

Suppose $$T : V \to W$$ and $$U$$ is a subspace of $$V$$ and is invariant under $$T$$. Then the map $$T' : V/U \to W$$ defined by $$T'(v + U) = Tv$$ is a linear map. I am struggling to show that this map is well-defined. If $$U = \ker(T)$$, then it would be easy to show. However, since $$U$$ is just any invariant subspace, I'm not sure how to proceed.

I started by saying suppose $$v_1 + U = v_2 + U \implies v_1 - v_2 \in U$$. Since $$U$$ is only invariant, and we made no assumption on its relation to the kernel of $$T$$, I can't say $$T(v_1 - v_2) = 0 \implies Tv_1 = Tv_2$$.

• "$U$ is invariant under $T$" makes no sense when $W\neq V$. "$T$-invariant" means $T(U)\leq U$, so it usually only applies to operators. What are you really doing? Commented Oct 15, 2022 at 20:41
• This doesn't work in general. Take $T\colon \mathbb{R}^2\to\mathbb{R}^2$ by $T(x,y)=(0,y)$. Then $U=\{(0,y)\in\mathbb{R}^2\mid y\in \mathbb{R}\} = R(T)$ is $T$-invariant. If we tried to define $T\colon \mathbb{R}^2/U\to\mathbb{R}^2$ by $T(v+U) = T(v)$, this is not well defined: $(2,3)+U = (2,4)+U$ since $(2,4)-(2,3)=(0,1)\in U$, but $T(2,3)=(0,3)$ and $T(2,4) = (0,4)\neq T(2,3)$. Commented Oct 15, 2022 at 20:48
• Indeed, the statement is just false. What you need is to assume that $U\subseteq Ker(T)$.
– Mark
Commented Oct 15, 2022 at 20:51
• In general, the map $T\colon V/U\to W$ given by $T(v+U)=T(v)$ is well-defined if and only if $U\leq \ker(T)$. Indeed, we know that if $U\leq \ker(T)$ then this is well defined. If $U$ is not contained in $\ker(T)$, then let $u\in U$ with $u\notin \ker(T)$. Then $0+U=u+U$, but $T(0)\neq T(u)$. Commented Oct 15, 2022 at 20:53

As @Arturo noted the question is ill formed if $$V \neq W$$ (or at least $$U \subseteq W$$). Invariance means $$T(U) \subseteq U$$ and since $$T(U) \subseteq W$$ the image must be of the same space.
But even if we were to assume $$V = W$$, this statement still wouldn't be true.
Consider $$T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ given by $$\pmatrix{5& 0\\ 0& 5}$$, very simple. Now of course the $$x$$-axis is an invariant subspace, call it $$U$$.
But notice $$(1, 7) + U = (2, 7) + U$$ but obviously $$T(2, 7) = (10, 35) \neq (5, 35) = T(1, 7)$$.
It would actually be a nice exercise to determine what kind of space $$U$$ must be in order to ensure well formedness. Hint, it should involve the kernel in some way after all.