Abstract Nonsense

Crushing one theorem at a time

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 - Posted by | Fun Problems, Halmos, Linear Algebra, Uncategorized | , , , ,

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