Endomorphisms and similarity of matrices
A linear mapping is called an endomorphism. We will be dealing with spaces which are finite dimensional. Thus, given a basis of we consider the matrix of the mapping in this basis . Notice that, if is also a basis of , then
This motivates the notion of similarity of matrices. We shall say that two matrices and are similar, if there exists an invertible matrix , such that . In other words, matrices and are similar if and only if they are matrices of the same endomorphism in two bases and .
It is easy to notice that the relation of similarity is an equivalence relation.
Invariants of similarity of matrices
If matrices and are similar, then it is easy to check that:
- the determinants of and are equal,
- sums of elements of diagonal (called the trace, denoted by ) are equal,
- ranks of matrices and are equal.
This justifies the possibility of talking about the determinant, the trace and the rank of an endomorphism, since those are the same for its matrix in any basis.
Idea: eigenvalues and eigenspaces
If you have tried to imagine a linear map of a plane, you usually imagine that it stretches or squeezes the plane along some directions. It would be nice to know whether a given linear map (such a map, which the same domain and range, is called an endomorphism), it is actually of this type. In other words we would like to know whether there exists a non-zero vector and a scalar , such that simply multiples by (so it stretches or squeezes the space in the direction of ), so:
If and have such properties, then is said to be an eigenvector of and an eigenvalue.
Notice that if is an eigenvalue of a map and is its eigenvector, then . Therefore if is a matrix of (e.g. in the standard basis, it actually does not matter), then
, where is the identity matrix.
Since multiplication of a matrix by a vector gives a linear combination of its columns and is a non-zero vector, we see that the columns of can be non-trivially combined to get the zero vector! It is possible if and only if . The expression is called the characteristic polynomial of . Notice that it does not depend on the basis in which the matrix is given.
How to find the eigenvalues of a map? Simply one needs to solve the equation . E.g let . Then:
So we have to solve the following:
And therefore the eigenvalues are: and .
Now let’s find eigenvectors related to subsequent eigenvalues. Notice that since is a linear map, if are eigenvectors for an eigenvalue , then for any scalar also and are eigenvectors for . Therefore, the set of all eigenvectors for forms a linear subspace. Notice that satisfies the equation
so the space of eigenvectors (i.e. eigenspace) for (denoted as ) is given by the following system of equations:
and we can easily find its basis.
In our example, let us find a basis of , so let . Then:
Therefore, we have the following system of equations:
The space of solutions is , and its basis is . Indeed, and .
Let’s find a basis of , so let . Then:
The system of equations:
In the reduced ,,stair-like” form:
The space of solutions is , and its basis is . Indeed, .
If the sum of dimensions of spaces related to the eigenvalues of a given map equals the dimension of the whole space (as in our example: ), then the basis of the whole space which consists of the vectors from the bases of subspaces related to the eigenvalues is called an eigenvector basis (in our case: ).
If a map has an eigenvector basis, then it can be actually described by means of squeezing and stretching in the directions of eigenvectors. Notice that the matrix of such a map in an eigenvector basis is an diagonal matrix (has non-zero elements only on its diagonal) with eigenvalues related to subsequent eigenvectors on its diagonal. In our example:
It may happen that a map has no eigenvectors (e.g. a rotation of the plane) or that the subspaces of eigenvectors are to small (e.g. a 10 degree rotation of a three-dimensional space around an axis had only one-dimensional space of eigenvectors).
Diagonalization of a matrix
A matrix is diagonalizable, if there exists a matrix , such that:
where is a diagonal matrix.
How to check it and diagonalize a matrix if it is possible? Simply consider a linear map such that is its matrix in standard basis. Matrix is diagonalizable, if and only if has eigenvector basis . Then:
and , since .
E.g. we know that
is diagonalizable, since related to this matrix has an eigenvector basis. furthermore in this case:
Calculating diagonalizable matrices
Diagonalization of a matrix has the following interesting matrix. We can use it to calculate a power of a matrix, if it is diagonalizable. Notice that if is a basis, then:
but if is a basis of eigenvectors, then is a diagonal matrix so calculating its power is simply calculating powers of the elements on the diagonal.
Let us show it on an example. Let us calculate: