# Abstract Nonsense

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

Point 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 book.

3.

Problem: The concept of a “linear transformation” as defined in the text, is too special for some purposes. According to a more general definition, a linear transformation from a vector space $\mathscr{U}$ to a vector space $\mathscr{V}$ over the same field is a correspondence $T$ that assigns a vector $u\in\mathscr{U}$ to a vector $T(u)\in\mathscr{V}$ in such a way that

$\displaystyle T\left(\alpha u+\beta u'\right)=\alpha T(u)+\beta T(u')$

Prove that each of the correspondences described below is a linear transformation in this generalized sense

a) $\varphi\in\text{Hom}\left(\mathscr{V},F\right)$ where $F$ is thought of as a one-dimensional space over itself.

b) $\pi_1:\mathscr{U}\boxplus\mathscr{V}\to\mathscr{U}:(u,v)\mapsto u$

Remark: This is usually referred to as the first canonical projection.

c) If $\mathscr{U}$ is a subspace of $\mathscr{V}$ then $T:\mathscr{V}\to\mathscr{V}/\mathscr{U}:v\mapsto v+\mathscr{U}$

d) Fix $B\in\text{Bil}\left(\mathscr{U},\mathscr{V}\right)$ and consider

$T_B:\mathscr{U}\to\text{Hom}\left(\mathscr{V},F\right):u_0\mapsto \varphi_{u_0}$

where $\varphi_{u_0}(v)=B(u_0,v)$.

Proof:

a) This is by definition since $\varphi(\alpha x+\beta y)=\alpha\varphi(x)+\beta \varphi(y)$.

b) This is fairly straight forward since

\begin{aligned}\pi_1\left(\alpha (u_1,v_1)+\beta (u_2,v_2)\right) &=\pi_1\left((\alpha u_1+\beta u_2,\alpha v_1+\beta v_2)\right)\\ &=\alpha u_1+\beta u_2\\ &=\alpha \pi\left((u_1,v_1)\right)+\beta \pi_1\left((u_2,v_2)\right)\end{aligned}

c) We merely note that by definition

\begin{aligned}T\left(\alpha v+\beta v'\right) &= \left(\alpha v+\beta v'\right)+\mathscr{U}\\ &= \alpha\left( v+\mathscr{U}\right)+\beta\left(v'+\mathscr{U}\right)\\ &= \alpha T(v)+\beta T(v')\end{aligned}

d) We merely recall that a bilinear form is linear in each slot if the other slot is fixed, and thus

\begin{aligned}\left(T_B(\alpha u+\beta u')\right)(v) &=B(\alpha u+\beta u',v)\\ &=\alpha B(u,v)+\beta B(u',v)\\ &=\alpha \left(T_B(u)\right)(v)+\left(T_B\left(u'\right)\right)(v)\end{aligned}

4.

a) Suppose that $\mathscr{U}$ and $\mathscr{V}$ are both $F$-spaces. If $AB$ are linear transformations from $\mathscr{U}$ to $\mathscr{V}$ and if $\alpha,\beta\in F$ and if

$C(x)=\alpha A(x)+\beta B(x)$

for each $x\in\mathscr{U}$ then $C$ is a linear transformation from $\mathscr{U}$ to $\mathscr{V}$.

b) If we write, by definition, $C=\alpha A+\beta B$, then the set of all linear transformations from $\mathscr{U}$ to $\mathscr{V}$ becomes a vector space with respect to this definition of linear operations.

c) Prove that if $\mathscr{U}$ and $\mathscr{V}$ are finite dimensional vector spaces, find the dimension of $\text{Hom}\left(\mathscr{U},\mathscr{V}\right)$.

Proof:

a) We merely note that

\begin{aligned}C(\alpha_0 x+\beta_0 y) &=\alpha A(\alpha_0 x+\beta_0 y)+\beta B(\alpha_0 x+\beta_0 y)\\ &=\alpha_0\left(\alpha A(x)+\beta B(x)\right)+\beta_0 \left(\alpha A(y)+\beta B(y)\right)\\ &=\alpha_0 C(x)+\beta_0 C(y)\end{aligned}

b) This is fairly routine, and is left to the reader.

c) For the definition of the terminology and a proof we refer to a previous post.

5.

Problem: Suppose that $\mathscr{M}$ is an $m$-dimensional subspace of a $n$-dimensional vector space $\mathscr{V}$. Prove that the set of those linear transformations $T$ on $\mathscr{V}$ for which $T\left(\mathscr{M}\right)=\{\bold{0}\}$ is a subspace of $\text{End}\left(\mathscr{V}\right)$. What is its dimension?

Proof: To prove that it’s a subspace we merely note that if $T,T'$ both annihilate $\mathscr{M}$, then for any $x\in\mathscr{M}$ we have that

$(\alpha T+\beta T')(x)=\alpha T(x)+\beta T'(x)=\alpha \bold{0}+\beta \bold{0}=\bold{0}$

and so $\alpha T+\beta T'$ annihilates $\mathscr{M}$. Let us, perhaps frivolously extend our notation of the annihilator of a set in a dual space, to an arbitrary space, in particular let

$\text{Ann}\left(\mathscr{U},\mathscr{W}\right)=\left\{T\in\text{Hom}\left(\mathscr{U},\mathscr{W}\right):T\left(\mathscr{U}\right)=\{\bold{0}\}\subseteq \mathscr{W}\right\}$

we claim then that with this notation, $\dim_F\text{Ann}\left(\mathscr{M},\mathscr{V}\right)=n(n-m)$. To do this let $\{x_1,\cdots,x_m\}$ be a basis for $\mathscr{M}$ and extend it to a basis $\{x_1,\cdots,x_m,x_{m+1},\cdots,x_n\}$ for $\mathscr{V}$. We claim that

$\mathcal{B}=\left\{T_{i,j}:i\in[n]-[m]j\in[n]\right\}$

form a basis for $\text{Ann}\left(\mathscr{M},\mathscr{V}\right)$ where $T_{i,j}$ is as it was in previous discussions. It’s evidently linear independent, since it’s a subset of a linearly independent set. Now, to see that it spans $\text{Ann}\left(\mathscr{M},\mathscr{V}\right)$ we let $T\in\text{Ann}\left(\mathscr{V},\mathscr{M}\right)$ be arbitrary. Then, since $\left\{T_{i,j}:i,j\in[n]\right\}$ spans $\text{End}\left(\mathscr{V}\right)$ there exists $\alpha_{i,j}\in F,\text{ }i,j\in[n]$ for which

$\displaystyle T=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i,j}T_{i,j}$

But, upon plugging in $i_0\in\{1,\cdots,m\}$ we get that

$\displaystyle 0=T(v_{i_0})=\sum_{j=1}^{n}\alpha_{i_0,j}T_{i_0,j}$

and so $\alpha_{i_0,j}=0,\text{ }j=1,\cdots,n$ and since $i_0$ was arbitrary it follows that $\alpha_{i,j}=0$ for $i\in[m]$ and $j\in[n]$. Thus, $\mathcal{B}$ spans $\text{Ann}\left(\mathscr{M},\mathscr{V}\right)$ as desired. The conclusion follows by noticing that $|\mathcal{B}|=n(n-m)$.

References:

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

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November 22, 2010 -

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