线性变换不可逆，则既不单也不满

abababa

设$V$是$n$维线性空间，$\sigma$是$V$上的线性变换，若$\sigma$不可逆，则是否有：$\sigma$既不是单射也不是满射？
我觉得只要满足一个就行了，比如$\sigma$不是单射，则$\sigma$不可逆。或者$\sigma$不是满射，则$\sigma$不可逆。

hbghlyj

Let $A:U→V$ be a linear operator between finite-dimensional vector spaces $U$ and $V$.
(1) $\operatorname{rank}(A)≤\dim(V)$
(2) $\operatorname{rank}(A)+\operatorname{nullity}(A)=\dim(U)$
(3) $A$ is injective iff $\operatorname{nullity}(A)=0$
(4) $A$ is surjective iff $\operatorname{rank}(A)=\dim(V)$
When $\dim U=\dim V$, $A$ is injective $\iff A$ is surjective $\iff A$ is bijective

abababa

我明白这个怎么回事了，刚才一时没转过来。因为$V$是有限维空间，所以单射即满射也即双射，而不可逆就不是双射，因此即不是单射也不是满射。

hbghlyj

On p. 326 of Roman's Advanced Linear Algebra, 3rd edition, he says:
    It's easy to see that an isometry $τ:V→W$ is always injective, but need not be surjective, even if $V=W$.
math.stackexchange.com/questions/838894
Note that for any finite-dimensional vector space $V$, a transformation $\tau:V \to V$ will be injective if and only if it is surjective (by the dimension theorem).

With that in mind, let's take a suitable infinite-dimensional example.

Let $V$ be the space of infinite sequences $(x_1,x_2,\dots)$ with $x_i \in \mathbb{R}$ (or $x_i \in \mathbb{C}$, if you prefer) for $i \in \mathbb{N}$ for which $\sum_{i=1}^\infty |x_i|^2 < \infty$.  We define the norm on $V$ via the inner product $\langle x,y\rangle = \sum_{i=1}^\infty x_i \overline{y_i}$. Define
$$
\tau[(x_1,x_2,\dots)] = (0,x_1,x_2,\dots)
$$
This example is often referred to as the right-shift operator.  Note that it is indeed an isometry on $V$ that is injective, but not surjective.

hbghlyj

Injectivity implies surjectivity

In some circumstances, an injective (one-to-one) map is automatically surjective (onto). For example,

Set theory
An injective map between two finite sets with the same cardinality is surjective.

Linear algebra
An injective linear map between two finite dimensional vector spaces of the same dimension is surjective.

General topology
An injective continuous map between two finite dimensional connected compact manifolds of the same dimension is surjective.

Czhang271828

本质上, 有合成列的模都满足 单同态=满同态=同构, 等价地, 同态单射=同态满射=同态双射.

称 $A$-模 $M$ 有合成列当且仅当有降链
\[
0=M_0\subsetneqq M_1\subsetneqq\cdots \subsetneqq M_{n-1}\subsetneqq M_n = M,
\]
且每一 $M_{k+1}/M_k$ 是单 $A$-模. 此后熟知 Jordan–Hölder 定理等都可以用起来. 特别地, 维度为 $n$ 的 $k$-线性空间是长度为 $n$ 的有合成列的 $k$-自由模.

对 $M$ 的任意 $A$ 模自同态, 短正合列
\[
0\to \ker f\to A\to \mathrm{im\,}f\to 0
\]
表明 $\ker f$ 的长度 $+$ $\mathrm{im\,}f$ 的长度 $=$ $A$ 的长度, 从而 $f$ 是单同态等价于 $f$ 是满同态, 等价于 $f$ 是同构.

hbghlyj

Fredholm Theory - Yale Math
A key result for Fredholm operators is that injectivity implies surjectivity + bounded inverse, which we prove in the theorem below.

Theorem 7. Suppose that $T = I +A$ where $A$ is compact. If $T$ is injective, then is surjective and has a bounded inverse.