0
$\begingroup$

The question is in the title. The solution gives this: "No; the set of isomorphisms does not contain the zero map (unless the space is trivial)."

To be honest, I have no idea how to solve this. I define subspace to be a subset, and my initial thoughts here is that a mapping between a vector space V to itself that's also one to one and onto is just a subset of mappings between a vector space V to itself that's not necessarily one to one or onto. How exactly does the zero map come into play here? Why does the zero map need to be in the set of isomorphisms in order for them to be considered a subspace of the space L(V, V)?

$\endgroup$
1
  • 2
    $\begingroup$ “I define a subspace to be a subset”. That is not the definition of a subspace. You should look up the definition in your notes or textbook and do some exercises with it - it’s a critically important concept in linear algebra. $\endgroup$
    – Mark Saving
    May 30, 2022 at 22:07

1 Answer 1

Reset to default
1
$\begingroup$

I define subspace to be a subset

That's incorrect. A subspace of a vector space $V$ is a subset $W\subseteq V$ which is also a vector space in its own right (with the addition and scalar multiplication "coming from $V$" in the obvious way). This means that $W$ must be

  • closed under addition (for each $a,b\in W$ we also have $a+b\in W$),

  • closed under scalar multiplication (for each scalar $s$ and each $a\in W$ we also have $s \cdot a\in W$), and

  • contain the zero vector,

where "$+$," "$\cdot$," and "zero vector" all take their meaning from $V$.

This more nuanced meaning of "sub[thing]" is ubiquitous in mathematics; compare "subgroup," "subring," ...

For example, consider $\mathbb{R}^2$ as a vector space over $\mathbb{R}$ in the usual way. The subset $$\{(0,r):r\in\mathbb{R}\}$$ is a subspace of $\mathbb{R}^2$, but $\{(1,r): r\in\mathbb{R}\}$ is not (in fact the latter fails all three of the bulletpoints above). The set $L(V,V)$ of your question is more difficult to picture, but the point is the same: the "zero vector" of $L(V,V)$ is the constant map $${\bf v}\mapsto{\bf 0}_V,$$ but this is usually not an isomorphism from $V$ to itself.

Exercise: when is the zero map an isomorphism?

Isomorphisms have to be bijective, so the zero map $V\rightarrow V$ is an isomorphism only if $V$ has exactly one element. And it's easy to check that in this case the zero map is indeed an isomorphism.

$\endgroup$
3
  • $\begingroup$ I see; the part about $${\bf v}\mapsto{\bf 0}_V,$$ makes a lot of sense. Is there somewhere that I can learn about why this is the case? I haven't gotten too much practise dealing with functions as vector spaces in the past, and I'm afraid I don't see how a mapping can have a zero vector at all. $\endgroup$
    – Mezzoforte
    May 30, 2022 at 22:35
  • $\begingroup$ @Mezzoforte An individual function isn't a vector space; rather, an appropriate set of functions may be (construable in some natural way as) a vector space. The standard example is the set $\mathscr{C}$ of continuous functions from $\mathbb{R}$ to $\mathbb{R}$. This is a vector space over $\mathbb{R}$, with vector addition $\oplus$ and scalar multiplication $\odot$ given by $$f\oplus g=\lambda x.f(x)+g(x)\quad\mbox{and}\quad s\odot f=\lambda x. s\cdot f(x).$$ The zero vector in this space is then the constant function $$\lambda x.0$$ (the "zero function"), which maps every input to zero. $\endgroup$
    – Noah Schweber
    May 30, 2022 at 22:40
  • $\begingroup$ You should check that this $\mathscr{C}$ does in fact satisfy the conditions of vector-space-over-$\mathbb{R}$-ness. Also, you should check that it is really very big: it has no finite spanning set (= is an infinite dimensional vector space)! $\endgroup$
    – Noah Schweber
    May 30, 2022 at 22:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .