# Abstract Nonsense

## Halmos Section 36: Inverses (Pt. II)

Point of post: This is a continuation of this post.

3.

Problem: Let $\mathscr{V}$ be any (not necessarily finite dimensional) $F$-space and $A,B\in\text{End}\left(\mathscr{V}\right)$. Then, $A,B\in\text{GL}\left(\mathscr{V}\right)$ if and only if $AB,BA\in\text{GL}\left(\mathscr{V}\right)$

Proof: We first prove a set-theoretic lemma:

Lemma: Let $f:X\to Y,g:Y\to Z$ then if $g\circ f$ is injective then $f$ is injective. Similarly, if $g\circ f$ is surjective then $g$ is surjective.

Proof: Suppose first that $g\circ f$ is injective but $f$ isn’t. Then, there exists $x\ne y\in X$ such that $f(x)=f(y)$ then evidently $g(f(x))=g(f(y))$ which contradicts that $g\circ f$ is injective.

Similarly, suppose that $g\circ f$ is surjective but $g$ isn’t. We then see that $g(f(X))\subseteq g(Y)\subsetneq Z$ which contradicts that $g\circ f$ is surjective.

The conclusion follows. $\square$

Corollary: If $f,g:X\to X$ and $f\circ g$ and $g\circ f$ are bijective then $f$ and $g$ are bijective.

Proof: Since $g\circ f$ is injective we know that $f$ is injective and since $f\circ g$ is surjective we know that $f$ is surjective. Thus, $f$ is bijective.

Similarly, since $f\circ g$ is injective we know that $g$ is injective and since $g\circ f$ is surjective we know that $g$ is surjective. Thus, $g$ is bijective.

The conclusion follows. $\square$

From this it readily follows that $AB,BA\in\text{GL}\left(\mathscr{V}\right)$ implies that $A,B\in\text{GL}\left(\mathscr{V}\right)$.  The other direction was covered in our discussion of invertible transform.

4.

Problem: If $\mathscr{V}$ is a finite dimensional $F$-space and $A,B\in\text{End}\left(\mathscr{V}\right)$ prove that if $AB=\mathbf{1}$ then $A,B\in\text{GL}\left(\mathscr{V}\right)$.

Proof: You’ll recall from our post on invertible transformations that $AB=\mathbf{1}$ implies that $A$ has a right inverse and $B$ a left. Thus, $A$ is an epimorphism (since it’s surjective) and $B$ is a monomorphism (since it’s injective). But, from the same post we know that this implies, by the finite dimensionality of $\mathscr{V}$, that $A,B\in\text{GL}\left(\mathscr{V}\right)$.

5.

Problem: If $\mathscr{V}$ is an $F$-space and $A,B,C,D\in\text{End}\left(\mathscr{V}\right)$ and $A-B,A+B\in\text{GL}\left(\mathscr{V}\right)$ then there exists $X,Y\in\text{End}\left(\mathscr{V}\right)$ such that

$AX+BY=C$

and

$BX+AY=D$

Proof: Assuming that $\text{char }F>2$ we see note that if the above two equations hold then upon addition we obtain

$(A+B)X+(A+B)Y=C+D$

so that

$X+T=(C+D)(X+Y)^{-1}$

and upon subtraction we obtain

$(A-B)X-(A-B)Y=C-D$

so

$X-Y=(C-D)(A-B)^{-1}$

solving this for $X$ gives

$\displaystyle X=\frac{1}{2}\left[(C+D)(A+B)^{-1}+(C-D)(A-B)^{-1}\right]$

and solving for $Y$ gives

$\displaystyle Y=\frac{-1}{2}\left[(C+D)(A+B)^{-1}-(C-D)(A-B)^{-1}\right]$

from where the conclusion follows.

6.

Problem:

a) Prove that if $\mathscr{V}$ is an $n$-dimensional $F$-space and $T\in\text{End}\left(\mathscr{V}\right)$, then $T\in\text{GL}\left(\mathscr{V}\right)$ if and only if $\mathscr{X}$ is linearly independent implies $T\left(\mathscr{X}\right)$ is linearly independent.

b) Is this assumption of finite dimensionality needed here?

Proof:

a) This was proven, in some detail (recalling that an endomorphism on a finite dimensional space is an isomorphism if and only if it’s a monomorphism) here

b) Yes. Consider the unique endomorphism

$T:\mathbb{C}[x]\to\mathbb{C}[x]$

such that $T(x^k)=x^{k+1}$ for all $k\in\mathbb{N}\cup\{0\}$ we claim that this mapping is linear independence preserving.  To see this suppose

$\displaystyle \left\{\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x^{s_{1,j_1}},\cdots,\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x^{s_{n,j_n}}\right\}=\mathcal{X}$

is linearly independent. Suppose then that there existed scalars $\beta_1,\cdots,\beta_n$ such that

$\displaystyle \beta_1 T\left(\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x^{s_{1,j_1}}\right)+\cdots+\beta_n T\left(\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x^{s_{n,j_n}}\right)=\bold{0}$

this implies by definition that

$\displaystyle \bold{0}=\beta_1 \sum_{j_1=1}^{m_1}\alpha_{1,j_1}x^{s_{1,j_1}+1}+\cdots+\beta_n \sum_{j_n=1}^{m_n}\alpha_{n,j_n} x^{s_{n,j_n}+1}=x\left(\beta_1 \sum_{j_1=1}^{m_n}\alpha_{1,j_1}x^{s_{1,j_1}}+\cdots+\beta_n \sum_{j_n=1}^{m_n}\alpha_{n,j_n}x^{s_{n,j_n}}\right)$

but since $\mathbb{C}[x]$ is an integral domain this implies that

$\displaystyle \beta_1\sum_{j_1=1}^{m_1}\alpha_{1,j_1}x^{s_{1,j_1}}+\cdots+\beta_n\sum_{j_n=1}^{m_n}\alpha_{n,j_n}x^{s_{n,j_n}}=\bold{0}$

and thus $\beta_1=\cdots=\beta_n=0$ from where it follows that $T\left(\mathscr{X}\right)$ is linearly independent. Thus, since $\mathscr{X}$ was arbitrary it follows that $T$ is linear independence preserving. But, note that $1\notin T\left(\mathbb{C}[x]\right)$ and so $T$ isn’t an epimorphism and thus not invertible.

7.

Problem: Show that if $\mathscr{V}$ is a finite-dimensional $F$-space and $T\in\text{End}\left(\mathscr{V}\right)$ is such that

$T^2-T+\mathbf{1}=\mathbf{0}$

prove that $T\in\text{GL}\left(\mathscr{V}\right)$.

Proof: Evidently $T$ is injective, since if $x\in\ker T$ then

$\mathbf{0}=T(T(x))-T(x)+\mathbf{1}(x)=T(\mathbf{0})-\mathbf{0}+x=\mathbf{0}+x=x$

so that $\ker T=\{\bold{0}\}$. But, by a previous theorem we know that $T\in\text{GL}\left(\mathscr{V}\right)$.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print