Abstract Nonsense

Crushing one theorem at a time

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 - 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. II) « Abstract Nonsense | March 29, 2011 | Reply


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