What is $\tilde{T}$ in Axler's Linear Algebra?

Question

On page 97 in section 3.E Axler gives definition 3.90: "Suppose $T \in L(V,W)$. Define $\tilde{T}: V/($null $T) \to W$ by $\tilde{T}(v+$null $T)=Tv$." Immediately preceding this was just some theorems about the dimensions of quotient spaces.

I don't quite understand what the purpose $\tilde{T}$ is, as it doesn't seem to be present in any of the exercises (at least not for 3.E), and it's not expanded upon afterward aside from presenting some statements about it's properties (e.g. $V/($null $T)$ is isomorphic to range $T$). I don't even understand what the map "represents". Isn't $v+$null $T$ a set? But $\tilde{T}$ doesn't equal another set, but rather a particular vector, $Tv$.

Does anyone have insight to this? Or a more common name for $\tilde{T}$? I haven't been able to find much on it anywhere.

Answer 1

$\tilde{T}$ is a very common construction... in abstract algebra. Not so much in linear algebra per se, but it fits into the general framework. No doubt this is why Axler introduces it, but does not do much with it.

Specifically: as you probably know already, given a vector space $V$ and a subspace $X$ of $V$, you can define the equivalence relation "congruent modulo $X$", by $$\mathbf{v}_1\sim \mathbf{v}_2 \iff \mathbf{v}_1-\mathbf{v}_2 \in X.$$

This equivalence relation partitions $V$ into equivalence classes, and it is a good exercise to verify that the equivalence class of $\mathbf{v}\in V$ modulo $X$ is given by $$[\mathbf{v}] = \{ \mathbf{v}+\mathbf{x}\mid \mathbf{x}\in X\}$$ and this set is denoted $\mathbf{v}+X$.

The set of equivalence classes is denote $V/X$, and it turns out that this set can be given a natural structure of a vector space (with vector addition $+_X$ and scalar multiplication $\cdot_X$), induced by the vector space structure of $V$, by defining $$\begin{align*} (\mathbf{v}_1+X) +_X (\mathbf{v}_2 + X) &= (\mathbf{v}_1+\mathbf{v}_2) + X\\ \alpha\cdot_X(\mathbf{v}+X) &= (\alpha\mathbf{v}) + X. \end{align*}$$

Now, given a linear transformation $T\colon V\to W$, we can associate to this linear transformation two natural subspaces: one of $V$ and one of $W$. The subspace of $V$ is $\mathrm{null}(T)$; the subspace of $W$ is $\mathrm{Im}(T)$. These are connected via the following important theorem, which is the vector space special case of what is often called the First Isomorphism Theorem of algebras (in the sense of universal algebra):

Theorem. Let $V$ and $W$ be vector spaces, and let $T\colon V\to W$ be a linear transformation. Then $V/\mathrm{null}(T)$ is isomorphic to $\mathrm{Im}(T)$. In fact, the linear transformation $\tilde{T}\colon V/\mathrm{null}(T)\to W$ defined by $$\tilde{T}(\mathbf{v}+\mathrm{null}(T)) = T(v)$$ is one-to-one, with image $\mathrm{Im}(T)$, and so $\tilde{T}$ induces the isomorphism.

That is $\tilde{T}$ is a function; it's defined on the quotient $V/\mathrm{null}(T)$. Turns out, it's a one-to-one linear transformation induced (closely connected to) $T$.

There are lots of pieces to verify in this theorem:

1. You must verify that $\tilde{T}$ is well-defined. Note that it is defined in terms of the "name"/representative of the elements of $V/\mathrm{null}(T)$; but each element there has many "names". So you would have to verify that if $\mathbf{v}_1+\mathrm{null}(T) = \mathbf{v}_2+\mathrm{null}(T)$ (if they are the same element of $V/\mathrm{null}(T)$), then $\tilde(T)(\mathbf{v}_1+\mathrm{null}(T)) = T(\mathbf{v}_1)$ is in fact equal to $\tilde(T)(\mathbf{v}_2 + \mathrm{null}(T)) = T(\mathbf{v}_2)$.

2. You must verify that $\tilde{T}$ is a linear transformation.

3. You must verify that $\tilde{T}$ is one-to-one.

4. You must verify that $\mathrm{Im}(T) = \mathrm{Im}(\tilde{T})$.

As I mention above, $\tilde{T}$ is a special case of a general construction. In a way, it represents the "essence" of $T$: because the value of $T$ on any given vector $\mathbf{v}\in V$ depends only on its equivalence class modulo $\mathrm{null}(T)$.

For example: if you have a linear transformation $T\colon\mathbb{R}^2\to\mathbb{R}^2$ given by $T(a,b) = (a,a)$, then $\mathrm{null}(T)=\{(0,b)\mid b\in\mathbb{R}\}$. We then have that $(a,b)\sim (c,d)\iff a=c$. And so we see that the equivalence class of $(a,b)\in \mathbb{R}^2$ depends only on the first coordinate... which is also the only thing that the value of $T$ depends on. In this example, the function $\tilde{T}\colon \mathbb{R}^2/\mathrm{null}(T)\to\mathbb{R}^2$ is defined by $$ \tilde{T}\Bigl( (a,b)+\mathrm{null}(T)\Bigr) = T(a,b) = (a,a).$$

In linear algebra this theorem is not very powerful, because we have a very easy way to check isomorphism: look at the dimension. Since we know that $\dim(X) + \dim(V/X) = \dim(V)$, and we know that $\dim(V) = \mathrm{nullity}(T)+\mathrm{rank}(T)$, this automatically tells us that $\mathrm{rank}(T)=\dim(V/X)$, and so $\mathrm{rank}(\tilde(T)) = \mathrm{rank}(T)$ if and only if $\mathrm{nullity}(\tilde(T)) = 0$.

In group theory, ring theory, and other areas of abstract algebra, this is a very useful theorem (hence its name as the First Isomorphism Theorem).

In summary: not much use right now for you, but a special case of something that will be very useful later on. Axler is introducing you to it now so that you have an example to think about when you encounter it later in your studies of abstract algebra.