# Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 25

*The dimension formula*

The following statement is called *dimension formula*.

Let denote a field, let and denote -vector spaces, and let

denote a -linear mapping. Suppose that has finite dimension. Then

holds.

### Proof

Let denote a field, let and denote -vector spaces, and let

denote a -linear mapping. Suppose that has finite dimension. Then we call

*rank*of .

The dimension formula can also be expressed as

We consider the linear mapping

given by the matrix

To determine the kernel, we have to solve the homogeneous linear system

The solution space is

and this is the kernel of . The kernel has dimension one, therefore the dimension of the image is , due to the dimension formula.

Let denote a field, let and denote -vector spaces with the same dimension . Let

denote a linear mapping. Then is injective if and only if is surjective.

This follows from the dimension formula and Lemma 24.14 .

*Composition of linear mappings and matrices*

In the correspondence between linear mappings and matrices, the composition of linear mappings corresponds to the matrix multiplication. More precisely: let denote vector spaces over a field with bases

Let

denote linear mappings. Then, for the describing matrix of , and of the composition , the relation

holds.

We consider the chain of mappings

Suppose that is described by the -matrix , and that is described by the -matrix (with respect to the bases). The composition has the following effect on the base vector .

Here, these coefficients are just the entries of the product matrix .

From this, we can conclude that the product of matrices is associative.

*Invertible matrices*

Let denote a field. For an invertible matrix , the matrix fulfilling

is called the *inverse matrix* of . It is denoted by

*Linear mappings and change of basis*

Let denote a field, and let and denote finite-dimensional -vector spaces. Let and be bases of and and bases of . Let

denote a linear mapping, which is described by the matrix with respect to the bases and . Then is described with respect to the bases and by the matrix

where and are the transformation matrices, which describe the change of basis from to and from to .

### Proof

Let denote a field, and let denote a -vector space of finite dimension. Let

be a linear mapping. Let and denote bases of . Then the matrices which describe the linear mapping with respect to and respectively (on both sides), fulfil the relation

This follows directly from Lemma 25.8 .

Two square matrices
are called
*similar*,
if there exists an
invertible matrix
with

Due to Corollary 25.9 , for a linear mapping , the describing matrices with respect to several bases are similar.

*Properties of linear mappings*

Let be a field, and let and be vector spaces over of dimensions and . Let

be a linear map, described by the matrix

with respect to two bases. Then the following properties hold.- is injective if and only if the columns of the matrix are linearly independent.
- is surjective if and only if the columns of the matrix form a generating system of .
- Let . Then is bijective if and only if the columns of the matrix form a basis of , and this holds if and only if is invertible.

Let and denote the bases of and respectively, and let denote the column vectors of . (1). The mapping has the property

where is the -th entry of the -th column vector. Therefore,

This is if and only if for all , and this is equivalent with

For this vector equation, there exists a nontrivial tuple , if and only if the columns are linearly dependent, and this holds if and only if is not injective.

(2). See
Exercise 25.3
.

(3). Let
.
The first equivalence follows from (1) and (2). If is bijective, then there exists a
(linear)
inverse mapping
with

Let denote the matrix for , and the matrix for . The matrix for the identity is the identity matrix. Because of Lemma 25.5 , we have

and therefore is invertible. The reverse implication is proved similarly.

*Finding the inverse matrix*

Let denote a square matrix. How can we decide whether the matrix is invertible, and how can we find the inverse matrix ?

For this we write down a table, on the left-hand side we write down the matrix , and on the right-hand side we write down the identity matrix
(of the right size).
Now we apply on both sides step by step the same elementary row manipulations. The goal is to produce in the left-hand column, starting with the matrix, in the end the identity matrix. This is possible if and only if the matrix is invertible. We claim that we produce, by this method, in the right column the matrix in the end. This rests on the following *invariance principle*. Every elementary row manipulation can be realized as a matrix multiplication with some
elementary matrix
from the left. If in the table we have somewhere the pair

after the next step (in the next line) we have

If we multiply the inverse of the second matrix (which we do not know yet, but we do know its existence, in case the matrix is invertible) with the first matrix, then we get

This means that this expression is not changed in each single step. In the beginning, this expression equals , hence in the end, the pair must fulfil

We want to find for the matrix its inverse matrix , following Method 25.12 .

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