## 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 to a vector space over the same field is a correspondence that assigns a vector to a vector in such a way that

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

**a) ** where is thought of as a one-dimensional space over itself.

**b) **

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

**c) **If is a subspace of then

**d) **Fix and consider

where .

**Proof:**

**a) **This is by definition since .

**b) **This is fairly straight forward since

**c) **We merely note that by definition

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

4.

**a) **Suppose that and are both -spaces. If are linear transformations from to and if and if

for each then is a linear transformation from to .

**b) **If we write, by definition, , then the set of all linear transformations from to becomes a vector space with respect to this definition of linear operations.

**c) **Prove that if and are finite dimensional vector spaces, find the dimension of .

**Proof:**

**a) **We merely note that

**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 is an -dimensional subspace of a -dimensional vector space . Prove that the set of those linear transformations on for which is a subspace of . What is its dimension?

**Proof: **To prove that it’s a subspace we merely note that if both annihilate , then for any we have that

and so annihilates . Let us, perhaps frivolously extend our notation of the annihilator of a set in a dual space, to an arbitrary space, in particular let

we claim then that with this notation, . To do this let be a basis for and extend it to a basis for . We claim that

form a basis for where 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 we let be arbitrary. Then, since spans there exists for which

But, upon plugging in we get that

and so and since was arbitrary it follows that for and . Thus, spans as desired. The conclusion follows by noticing that .

**References:**

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

No comments yet.

## Leave a Reply