Abstract Nonsense

Halmos Section 36: Inverses (Pt. I)

Point of post: In this post we complete the problems at the end of section 36 in Halmos.

1.

Problem: Which of the following linear transformations are invertible?

a) $T:\mathbb{C}\to\mathbb{C}:z\mapsto \overline{z}$, where $\mathbb{C}$ is considered a real vector space.

b) $T:\mathbb{C}[x]\to\mathbb{C}[x]$ where $\left(T(p)\right)(x)=p(x+1)-p(x)$.

c) $T:\mathscr{V}^{\otimes k}\to\mathscr{V}^{\otimes k}:x_1\otimes\cdots\otimes x_k\mapsto x_{\pi(1)}\otimes\cdots\otimes x_{\pi(k)}$, where we clearly this is meant to mean that $T$ is the unique element of $\text{End}\left(\mathscr{V}^{\otimes k}\right)$ for which $x_1\otimes\cdots\otimes x_k\mapsto x_{\pi(1)}\otimes\cdots\otimes x_{\pi(k)}$

d) $T:\text{Mult}_k\left(\mathscr{V}\right)\to\text{Mult}_k\left(\mathscr{V}\right):K\mapsto\pi K$

e) $T:\text{Mult}_k\left(\mathscr{V}\right)\to\text{Mult}_k\left(\mathscr{V}\right):K\mapsto \text{Sym}\left(K\right)$

f) $T:\text{Mult}_k\left(\mathscr{V}\right)\to\text{Mult}_k\left(\mathscr{V}\right):K\mapsto\text{Asym}\left(K\right)$

Proof:

a) Since $\dim_\mathbb{R} \mathbb{C}=2$ it suffices to show that $T$ is injective. But, this is clear since

$x-iy=x'-iy'\implies x=x'\text{ and }-y=-y'\implies x+iy=x'+iy'$

and so injectivity follows. In fact, a quick computation shows that $T^{-1}=T$.

b) $T\notin\text{GL}\left(\mathbb{C}[x]\right)$ since $1\in\ker T$

c) We merely recall that it’s sufficient to check that $T$ injects a basis into a basis. Recall though that

$\left\{x_{j_1}\otimes\cdots\otimes x_{j_k}:j_1,\cdots,j_k\in[k]\right\}=\mathscr{B}$

is a basis for $\mathscr{V}^{\otimes k}$ and evidently $T:\mathscr{B}\to\mathscr{B}$ is an injection and so $T$ is a monomorphism. But, since $\mathscr{V}^{\otimes k}$ is finite dimensional the conclusion follows.

d) Suppose that $\pi K=\bold{0}$ and let $(x_1,\cdots,x_k)\in\mathscr{V}^k$ be arbitrary. We then see that

$0=\left(\pi K\right)(x_{\pi(1)},\cdots,x_{\pi(k)})=K(x_{\pi^{-1}(\pi(1))},\cdots,x_{\pi^{-1}(\pi(k)})=K(x_1,\cdots,x_k)$

But, since $(x_1,\cdots,x_k)$ was arbitrary it follows that $K=\bold{0}$. Thus, $\ker T=\{\bold{0}\}$ and so $T$ is a monomorphism. Since $\text{Mult}_k\left(\mathscr{V}\right)$ is finite dimensional we may conclude that $T$ is invertible.

e) This is not invertible because it isn’t injective. Recall that for a fixed $\pi\in S_k$ $f:S_k\to S_k:\sigma\mapsto \sigma\pi$ is a bijection. Thus,

$\displaystyle \text{Sym}\left(K\right)=\sum_{\sigma\in S_k}\sigma K=\sum_{sigma \in S_k}\pi\sigma \left(\pi^{-1}K\right)=\pi^{-1}\sum_{\sigma\in S_k}\sigma \left(\pi^{-1}K\right)=\text{Sym}\left(\pi^{-1}K\right)$

Thus, choosing a non-symmetric $K$ and using the above shows that $T$ is not injective and thus not invertible.

f) The same concept applies

$\displaystyle \text{Asym}\left(K\right)=\sum_{\sigma \in S)k}\text{sgn}(\sigma)\sigma K=\text{sgn}(\pi^{-1})\sum_{\sigma\in S_k}\text{sgn}(\sigma\pi)\sigma\pi\left(\pi^{-1}K\right)=\text{sgn}(\pi^{-1})\sum_{\sigma\in S_k}\text{sgn}(\sigma)\sigma\left(\pi^{-1}K\right)=\text{sgn}(\pi^{-1}) \text{Asym}\left(\pi^{-1}K\right)$

so choosing an even $\pi$ and a suitable $K$ the above violates injectivity. Thus, $T$ is not injective thus not invertible.

2.

Problem: Consider $T\in\text{End}\left(\mathbb{C}^2\right)$ given by

$T:\left(\zeta_1,\zeta_2\right)\mapsto \left(\alpha \zeta_1+\beta\zeta_2,\gamma\zeta_1+\delta\zeta_2\right)$

Prove that $T\in\text{GL}\left(\mathbb{C}^2\right)$ if and only if $\alpha\delta-\beta\gamma\ne 0$.

Proof: Suppose that $\alpha\delta-\beta\gamma=0$. If $(\delta,\gamma)=(0,0)$ we’re done since the equation $\alpha\delta-\beta\gamma=0$ is trivially satisfied and $T((0,1))=T((0,0))=(0,0)$, and thus $T$ is not injective. So, assume not. Note then that

\begin{aligned}(0,0) &= (\alpha\delta-\beta\gamma,\delta\gamma-\delta\gamma)\\ &= \delta\left(\alpha,\gamma\right)-\gamma\left(\beta,\delta\right)\\ &= \delta T\left((1,0)\right)-\gamma T((0,1))\\ &= T\left((\delta,-\gamma)\right)\end{aligned}

and since $(\delta,\gamma)\ne (0,0)$ it follows that $\ker T\ne\left\{(0,0)\right\}$ and so $T$ is not injective and thus not invertible. Thus, if $T\in\text{GL}\left(\mathscr{V}\right)$ then $\alpha\delta-\beta\gamma\ne 0$.

Conversely, suppose that $\alpha\delta-\beta\gamma\ne 0$. Now, clearly we can’t have $\beta=0$ and $\delta=0$ otherwise $\alpha\delta-\beta\gamma=0-0=0$ contradictory to our assumption. So, suppose that $(\zeta_1,\zeta_2)\in \ker T$ but $(\zeta_1,\zeta_2)\ne 0$. Then, we may assume without loss of generality that $\zeta_1\ne 0$. So, by assumption

$T\left((\zeta_1,\zeta_2)\right)=\left(\alpha\zeta_1+\beta\zeta_2,\gamma\zeta_1+\delta\zeta_2\right)=(0,0)$

and thus

$\alpha\zeta_1+\beta\zeta_2=\gamma\zeta_1+\delta\zeta_2=0$

We see from $\alpha\zeta_1+\beta\zeta_2=0$ that

$\displaystyle \frac{\alpha}{\beta}=\frac{-\zeta_2}{\zeta_1}\quad\mathbf{(1)}$

But, note that

$\gamma\zeta_1+\delta\zeta_2=0\implies \gamma+\delta\frac{\zeta_2}{\zeta_1}=0$

and so plugging $\mathbf{(1)}$ into this gives

$\displaystyle \gamma+\delta\left(\frac{-\alpha}{\beta}\right)=0$

and so

$\beta\gamma-\alpha\delta=0$

contradictory to our assumption. Thus, it follows that $\ker T=\left\{(0,0)\right\}$ and thus $T$ is injective. Therefore, since $\mathbb{C}^2$ is finite dimensional we may conclude that $T\in\text{GL}\left(\mathbb{C}^2\right)$.

References:

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

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December 2, 2010 -

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