# Abstract Nonsense

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

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

8.

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

$A(BA+B-\mathbf{1})=ABA+AB-A=A+\mathbf{1}-A=\mathbf{1}$

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 $B=A^{-1}$.

9.

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).

10.

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

$T_B:\mathscr{W}\to\text{Hom}\left(\mathscr{V},F\right):w\mapsto\varphi_w$

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).

References:

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