# Problems from Chapter 4: Linear Operations, Section 1: The Dimension Formula

## Problem 4.1.1

**Let be respectively matrices. Show that is a linear transformation **

Recall that a linear transformation is a map such that and where are vectors and is a scalar constant.

First note that the map is defined only when that is when it has m rows and n columns.

Then observe that by linearity of matrices.

Let then by distributivity of matrices.

So the map is linear.

A comment that this is really a very trivial problem and just follows from the composition of matrices being a linear map, however the question is an intuition testing question and so requires a more detailed answer.

## Problem 4.1.2

**Let be elements of a vector space . Prove that defined by**

**is a linear transformation.**

From the definitions:

and

Therefore the map is linear.

## Problem 4.1.3

**Let be an matrix. Use the dimension formula to prove that the space of solutions of the linear system has dimension at least .**

The dimension formula tells us

.

Here the kernel of are the solutions to by definition of the kernel.

A matrix is a linear map . With

Therefore , then by dimension formula

## Problem 4.1.4

**Prove that every matrix of rank 1 has the form , where are dimensional column vectors. How uniquely determined are these vectors?**

We need to show that every has rank one, and that every matrix has the form .

Let and

Then the row of is

Each row is in the span of therefore has rank one. It is useful to think of this as ‘the codomain of has dimension one’.

Let be an arbitrary rank 1 matrix. Then it’s rows are linear combinations of some vector . Therefore each row is of the form

If we assign the , into a vector . Then we get that

This construction depends on a choice of , which then determines . This choice could be any scalar multiple of a known . Thus the choice is not very unique at all, it is any real number other than zero.

There is however a canonical choice, which is

that is the ‘normal’ or ‘orthonormal’ element in the span of .

## Problem 4.1.5

**(a) ****Let be vector spaces over a field . Show that the two operations **

* and *

*make the product set into a product vector space.*

contains the zero element as . Closed under addition as closed under addition. Closed under scalar multiplication as closed under scalar multiplication.

**(b) Let be subspaces of a vector space . Show that **

** where **

**is a linear transformation.**

and

therefore transformation is linear.

**(c) Express the dimension formula for in terms of dimensions of subspaces of .**

Think of as being represented by it’s matrix which takes vectors in to vectors in .

Note that and not necessarily

Where

Therefore

and

This expresses the dimension formula for in terms of subspaces of . Noting that .