Abstract Nonsense

Crushing one theorem at a time

Halmos Section 36: Inverses (Pt. III)

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


Problem: If \mathscr{V} is an F-space and A\in\text{End}\left(\mathscr{V}\right). Prove that if A has precisely one right inverse then A is invertible.

Proof: We merely note that if AB=\mathbf{1} (taking B to be the unique right inverse) then


so that BA+B-\mathbf{1} is a right inverse for A, thus by assumption BA+B-\mathbf{1}=B and so solving this for BA gives BA=\mathbf{1}.  Thus A\in\text{GL}\left(\mathscr{V}\right) and B=A^{-1}.



Problem: If \mathscr{V} is a finite dimensional F-space and A\in\text{GL}\left(\mathscr{V}\right) then there exists a polynomial p such that p(A)=A^{-1}.

Proof: It’s true in general that given A\in\text{End}\left(\mathscr{V}\right) that there exists a minimal polynomial which annihilates this, but for the invertible case its’s easier to show that there exists just a polynomial with non-zero constant coefficient which annihilates A. Intuitively, since we know that there exists some non-zero polynomial

\displaystyle p(x)=\sum_{j=0}^{m}\alpha_j x^j

which annihilates A. If \alpha_0=0 then we may factor this as

\displaystyle A\left(\sum_{j=1}^{m}\alpha_j x^j\right)

but since A is invertible and the above expression identically zero we may conclude that

\displaystyle \sum_{j=1}^{m}\alpha_j A^{m-1}=\bold{0}

continuing with this process (realizing that it must clearly end otherwise we arrive at \alpha A is identically zero, which implies that A=\bold{0} which contradicts that A\in\text{GL}\left(\mathscr{V}\right)) we arrive at a polynomial

\displaystyle q(x)=\sum_{j=0}^{\ell}\alpha_j x^j

for which q(A)=\bold{0} and \alpha_0\ne 0. Thus, we see that

\displaystyle \alpha_\ell A^{\ell}+\cdots+\alpha_1 A=-\alpha_0\mathbf{1}

multiplying both sides by \displaystyle \frac{-1}{\alpha_0}A^{-1} proves that

\displaystyle \frac{-\alpha_\ell}{\alpha_0}A^{\ell-1}+\cdots+\frac{-\alpha_1}{\alpha_0}\mathbf{1}=A^{-1}

from where the conclusion follows.

Remark: Too the interested/precocious reader: you can attempt to prove the fact stated in the first sentence of the above paragraph by first considering the polynomial ring F[t] and and showing that \mathcal{I}_A=\left\{p\in F[t]:p(A)=\bold{0}\right\} is a proper ideal of F[t]. Prove then that \mathcal{I}_A is generated by some monic polynomial, and this polynomial is the one we seek).



Problem: Formulate a sensible definition of invertibility for linear transformations from one F-space \mathscr{V} to another F-space \mathscr{W}. Use that definition to determine which of the following linear transformations are invertible:

a) \varphi\in\text{Hom}\left(\mathscr{V},F\right)

b) \pi_1:\mathscr{V}\oplus\mathscr{W}\to\mathscr{V}:(v,w)\mapsto v

c) Assuming that \mathscr{W}\leqslant\mathscr{V} we may define T:\mathscr{V}\to\mathscr{V}/\mathscr{W}:v\mapsto v+\mathscr{W}

d) Fix B\in\text{Bil}\left(\mathscr{V},\mathscr{W}\right) and define


where \varphi_w(v)=B(v,w).

Proof: If T\in\text{Hom}\left(\mathscr{V},\mathscr{W}\right) we merely call T invertible if it’s bijective.

a) Recall from an earlier post that if \varphi\in\text{Hom}\left(\mathscr{V},F\right) and v_0\in\mathscr{V} is such that \varphi(v_0)\ne 0 then \mathscr{V}=\ker\varphi\oplus\text{span }\{v_0\}.In other words, \dim_F \ker T=\dim_F \mathscr{V}-1. Thus, recalling that T will be a monomorphism if and only if \ker T=\{\bold{0}\} which is true if and only if \dim_F\ker T=0 we may conclude that \varphi is an isomorphism (recalling that a non-zero linear functional is always surjective) if and only if \dim_F\mathscr{V}=1, in which case \mathscr{V}\cong F^{\dim_F\mathscr{V}}=F in which case, by an appropriate composition of maps, we can interpret \varphi as being an endomorphism.

b) This is evidently not a monomorphism (in most cases) since \ker\pi_1=\{\bold{0}\}\oplus\mathscr{W}. Thus, \pi_1 is invertible (since it’s evidently surjective) if and only if \dim_F\mathscr{W}=0 in which case \mathscr{V}\oplus\mathscr{W}\cong\mathscr{V} and the same idea of being able to view \pi_1 as an endomorphism discussed in a) still applies.

c) Note that if T were an isomorphism then \dim_F\mathscr{V}=\dim_F\mathscr{V}/\mathscr{W}=\dim_F\mathscr{V}-\dim\mathscr{W} and thus \dim_F\mathscr{W}=0. Thus, (as one can easily check the other direction) T is an isomorphism if and only if \mathscr{W}=\{\bold{0}\}.

d) Once again using a dimension argument we may conclude that the only time this could possibly be an isomorphism is when \mathscr{W}\cong\mathscr{V} (using a dimension argument). Moreover, this is highly dependent upon the identity of B. For example, if B=\bold{0} then this is evidently not an isomorphism. In fact, we can say this mapping is an isomorphism precisely when \mathscr{W}\cong\mathscr{V} and B is non-degenerate in the second column. In the sense that B(v,w_0)=\bold{0} for all v\in\mathscr{V} implies that w_0=\bold{0}. To see this we first note that if T_B(w)=T_B(w') then B(w,v)-B(w',v)=B(w-w',v)=0 for all v\in \mathscr{V}, which by assumption implies that w=w'. Thus, T_B is injective. Recalling though that since \mathscr{W}\cong\mathscr{V}\cong\text{Hom}\left(\mathscr{V},F\right) we may conclude that T_B is an isomorphism (since T_B(\mathscr{W}) is a subspace of \text{Hom}\left(\mathscr{V},F\right) with equal dimension).



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


December 4, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra | , , , , ,

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