Abstract Nonsense

Crushing one theorem at a time

Halmos Sections 32 and 33: Linear Transformations and Transformations as Vectors (Pt. I)

Point of post: In this post we complete the problems that appear at the end of Halmos, sections 32 and 33.


Problem: Prove that each of the correspondences described below is a linear transformation.

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]:p(x)\mapsto 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)}

d) T:\text{Mult}_k\left(\mathscr{V}\right)\to\text{Mult}_k\left(\mathscr{V}\right):K\mapsto \pi^{-1} K

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

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

Remark: If the notation seems unfamiliar, that’s because I created my notation. See here and here


a) We note that

\begin{aligned}T\left(\alpha_1 (a_1+bi)+\alpha_2(a_2+b_2 i)\right) &=T\left((\alpha_1 a_1+\alpha_2)+(\alpha_1 b_1+\alpha_2 b_2)\right)\\ &=\left(\alpha_1 a_1+\alpha_2 a_2\right)-(\alpha_1 b_1+\alpha_2 b_2)i\\ &=\alpha_1 a_1-\alpha_1 b_1+\alpha_2 a_2-\alpha_2 b_2 i\\ &=\alpha_1 T\left(a_1+b_1 i\right)+\alpha_2 T\left(a_2+b_2 i\right)\end{aligned}

Remark: It was crucial in the above that \alpha_1,\alpha_2\in\mathbb{R} since otherwise we might not be able to claim that the imaginary part was the sum of the original imaginary parts. For example, it’s not true that iT(1+2i)=T(i(1+2i)).

b) For this we merely note that

\begin{aligned}T(p+q) &=(p+q)(x+1)-(p+q)(x)\\ &=(p(x+1)+q(x+1))-(p(x)+q(x))\\ &=(p(x+1)-p(x))+(q(x+1)-q(x))\\ &=T(p)+T(q)\end{aligned}


T(\alpha p)(x)=(\alpha p)(x+1)-(\alpha p)(x)=\alpha(p(x+1)-p(x))=\alpha (T(p))(x)

from where the conclusion follows.

c) To do this merely note that for x_1\otimes\cdots\otimes x_k,x'_1\otimes\cdots\otimes x'_k\in \mathscr{V}^{\otimes k} we have that

\begin{aligned}T\left(\alpha(x_1\otimes\cdots\otimes x_k)\right) &=x_1\otimes\cdots\otimes (\alpha x_{\pi(1)})\otimes\cdots\otimes x_k\\ &=(\alpha x_{\pi(1)})\otimes\cdots\otimes x_{\pi(k)}\\ &=\alpha (x_{\pi(1)}\otimes\cdots\otimes x_{\pi(k)})\\ &=\alpha T\left(x_1\otimes\cdots\otimes x_k\right)\end{aligned}


\begin{aligned}\left(T\left(x_1\otimes\cdots\otimes x_k+x'_1\otimes\cdots\otimes x'_k\right)\right)\left(K\right) &= \left(x_{\pi(1)}\otimes\cdots\otimes x_{\pi(k)}+x'_{\pi(1)}\right)\left(K\right)\\ &= K\left(x_{\pi(1)},\cdots,x_{\pi(k)}\right)+K\left(x'_{\pi(1)},\cdots,x'_{\pi(k)}\right)\\ &= \left(T(x_1\otimes\cdots\otimes x_k)\right)(K)+\left(T(x'_1\otimes\cdots\otimes x'_k)\right)(K)\end{aligned}

And thus since every element of \mathscr{V}^{\otimes k} can be written as a linear combination of vectors of the form x_1\otimes\cdots\otimes x_k it follows that T is additive, from where the conclusion follows.

d) This is fairly similar, namely

\displaystyle \left(T(\alpha K)\right)(x_1,\cdots,x_k)=(\alpha K)(x_{\pi(1)},\cdots,x_{\pi(k)}=\alpha K(x_{\pi(1)},\cdots,x_{\pi(k)})=\alpha \left(T(K)\right)(x_1,\cdots,x_k)


\begin{aligned}\left(T(K+K')\right)(x_1,\cdots,x_k) &= \left(K+K'\right)(x_{\pi(1)},\cdots,x_{\pi(k)})\\ &= K(x_{\pi(1)},\cdots,x_{\pi(k)})+K'(x_{\pi(1)},\cdots,x_{\pi(k)})\\ &= \left(T(K)\right)(x_1,\cdots,x_k)+\left(T(K')\right)(x_1,\cdots,x_k)\end{aligned}

e) For this we merely note that by part d)

\displaystyle \begin{aligned}\left(T\left(\alpha K+\beta K'\right)\right)(x_1,\cdots,x_k) &= \sum_{\pi\in S_k}\left(\pi\left(\alpha K+\beta K'\right)\right)(x_1,\cdots,x_k)\\ &= \sum_{\pi\in S_k}\left(\alpha \left(\pi K\right)(x_1,\cdots,x_k)+\beta\left(\beta K'\right)(x_1,\cdots,x_k)\right)\\ &= \alpha \sum_{\pi\in S_k}\left(\pi K\right)(x_1,\cdots,x_k)+\beta \sum_{\pi\in S_k}\left(\pi K'\right)(x_1,\cdots,x_k)\\ &= \alpha \left(T(K)\right)(x_1,\cdots,x_k)+\beta \left(T(K')\right)(x_1,\cdots,x_k)\end{aligned}

f) Very similarly, using part d) again

\begin{aligned}\left(T\left(\alpha K+\beta K'\right)\right)(x_1,\cdots,x_k) &= \sum_{\pi\in S_k}\text{sgn}(\pi)\left(\pi\left(\alpha K+\beta K'\right)\right)(x_1,\cdots,x_k)\\ &=\sum_{\pi\in S_k}\left(\alpha \text{sgn}(\pi)\left(\pi K\right)(x_1,\cdots,x_k)+\beta\text{sgn}(\pi)\left(\pi K'\right)(x_1,\cdots,x_k)\right)\\ &=\alpha\sum_{\pi\in S_k}\text{sgn}(\pi)\left(\pi K\right)(x_1,\cdots,x_k)+\beta\sum_{\pi\in S_k}\text{sgn}(\pi)\left(\pi K'\right)(x_1,\cdots,x_k)\\ &= \alpha\left(T(K)\right)(x_1,\cdots,x_k)+\beta\left(T(K')\right)(x_1,\cdots,x_k)\end{aligned}

2. Problem: Prove that \text{End}\left(\mathscr{V}\right) is a vector space. What’s its dimension?

Proof: For the notation and proof see here


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


November 22, 2010 - Posted by | Fun Problems, Halmos, Linear Algebra | , , ,


  1. […] of post: This is a continuation of this post in an effort to answer the questions at the end of sections 32 and 33 in Halmos’s […]

    Pingback by Halmos Sections 32 and 33: Linear Transformations and Transformations as Vectors (Pt. II) « Abstract Nonsense | November 22, 2010 | Reply

  2. Can you clearly post solution for exercise 2 of this section? Thank you!

    Comment by Amira | November 29, 2010 | Reply

    • If you’ll kindly follow the link, I prove a more general statement there. Is that not satisfactory?

      Comment by drexel28 | November 29, 2010 | Reply

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