My question is in the context of linear algebra (of a linear map $T$ between two FDVS $V$ and $W$), though I’ve heard it is a more general result in group theory (First Isomorphism Theorem). I’ve seen some answers about intuition for this theorem, but they are in the context of group theory and more general/abstract.
Do the following observations related to this theorem hold? Am I thinking about this result correctly? I would also appreciate any general comments about this result (whether in linear algebra or in general).
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(Visualizing movement between points and between affine subsets) A movement between two points in the image corresponds to a movement between two affine subsets of the domain that are parallel to the kernel (and vice versa). (See image below (I drew this, so not sure if accurate).) Specifically, if we move from $T(x)\in\operatorname{im}T$ to $T(x’)\in\operatorname{im}T$ then we move from $x+\ker T$ to $x’+\ker T$ in the domain (and vice versa). In fact, we can be more general: we can use any element in the fiber of $T(x)$ to represent $x + \ker T$, and any element in the fiber of $T(x’)$ to represent $x’ + \ker T$.
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(Application to systems of linear equations) Let $A$ be the matrix of $T$. The solution set of $Ax = 0$ is $\ker T$. More generally, consider $Ax = b$. If this system is consistent, i.e. if $b$ is in the image of $T$, then its solution set is given by $v + \ker T$, where $v$ is any solution $Av = b$.
Edit: Here are a few more observations (some overlap a bit with what I’ve already said); they may or may not be accurate:
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(Removal of useless/trivial solutions) The map $T$ from $V$ to $W$ (or to im $T$) may not be injective, but if we think about the map $\tilde{T}$ from $V/\text{ker $T$}$ to im $T$ that sends $v + \text{ker $T$}$ to $v$, we get a bijection. We’ve sort of “removed duplicates” and achieved a one-to-one-correspondence. Also, we've collapsed the kernel into a single point (i.e. 0), so all the extra trivial solutions have been removed; now, only 0 gets mapped to 0.
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(Partitioning the domain into copies of the kernel) The quotient space $V/\text{ker $T$}$ is the set of all affine subsets parallel to the kernel. So intuitively the isomorphism tells us that if we chop up the domain into copies of the kernel then we get something isomorphic to the image.
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(Number of remaining directions required to fill up the domain) $V/\text{ker $T$}$ has dimension $\dim V – \dim \ker T$, so it sort of tells us how many more directions we need to fill up the space. E.g. in $\mathbb{R}^3$, if the kernel has dimension 2, then $V/\text{ker $T$}$ has dimension 1; there is “one more direction” to go. Geometrically, sweeping a plane through a line (that's not in the plane) gives us all of 3D space.
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(The image is isomorphic to the subspace perpendicular to the kernel) The orthogonal complement $U^\perp$ of a subspace $U$ has dimension $V – U$. Setting $U = \text{ker $T$}$ tells us that $(\ker T)^\perp$ has the same dimension as im $T$, i.e. they are isomorphic. Thinking in $\mathbb{R}^3$ again, if the kernel is a plane, then im $T$ and $(\ker T)^\perp$ are lines, so they are isomorphic. Thinking of $V/\ker T$ as chopping up $\mathbb{R}^3$ into a line of parallel planes also shows how $V/\ker T$ is like a line as well.
Best Answer
I favor the second interpretation, which I'll rewrite here:
Above every element $b \in \operatorname{im}(T)$, there is an affine subspace $a+\ker(T)$, where $T(a)=b$. Any $a$ such that $T(a)=b$ will do. Above here means the inverse image of $b$, that is, $T^{-1}(b)$, the fiber above $b$.
In other words, $T$ is a bit like a projection along $\ker(T)$, except that the codomain is $W$ not $V$.