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
Proof
Let denote a field, let and denote -vector spaces, and let
denote a -linear mapping. Suppose that has finite dimension. Then we call
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
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 that 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; however, 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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