Abstract Nonsense

Crushing one theorem at a time

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

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December 3, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , , ,

1 Comment »

  1. […] Point of post: This is a continuation of this post. […]

    Pingback by Halmos Section 36: Inverses (Pt. III) « Abstract Nonsense | December 4, 2010 | Reply


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