Linear algebra: Fifth chapter The eigenvalues of the inner product matrices of the similarity matrix and the two-second vector (1) and the similarity matrix _ linear algebra of eigenvectors

The inner product of the first section of a vector. Mathematical Concepts

1. Inner product: With n-dimensional vector

Make

It is called [X,y] as the inner product of the vector x and Y.

2. Norm: A norm (or length) called a vector x.

3. Unit vector: The vector x is called the unit vector.

4. When, when, said

is the angle between the vector x and Y.

5. Orthogonal vector group: refers to a group of 22 orthogonal unit vectors.

6. Standard orthogonal Base: The n-dimensional vector is a base of the vector space V, and if 22 is orthogonal and is a unit vector, it is called a standard orthogonal base of V.

7. Orthogonal matrix: If n-order phalanx a satisfies

Then the last is called a orthogonal matrix.

8. Orthogonal transformation: If P is an orthogonal matrix, then the linear transformation x=py is called an orthogonal transformation. Two Principles, formulas, and rules

1. The result of an inner product is a number (or a polynomial) that satisfies the following properties (where X,y,z is an n-dimensional vector, real numbers):

(i);

(ii);

(iii)

2. The norm of a vector is a number and satisfies the following properties:

(i) nonnegative when x≠0, when x = 0 o'clock,.

(ii) homogeneity;

(iii) triangular inequalities.

3. is the unit vector.

4. The orthogonal vector group is linearly independent.

5. Schmidt Standard Orthogonal

Set linearly independent,

Take, make

............................................................

Three. Analysis of key points and difficulties

This section mainly describes some preparatory knowledge, its focus is the vector of the inner product, norms, standard orthogonal matrix, orthogonal matrices and orthogonal transformation, will be enough to make orthogonal matrix, easy to get orthogonal transformation, orthogonal transformation is often used in the following learning, the difficulty is that the Schmidt standard orthogonal. Four. Typical examples

Example 1 known vectors, a set of nonzero vector a1,a2, which A1 and A2,a3 orthogonal, and a1,a2,a3 into a standard orthogonal base of R3.

Solution: Set the desired vector x, is [a3,x] = 0, i.e.

，

Its basic solution is

，

Make

, they are orthogonal to the standard,

，

，

， ,

is obviously a 22 orthogonal unit vector, so it is a standard orthogonal base.

The key of the Jiuben problem is orthogonal with the vector and the known vector, and the inner product is equal to zero, and the homogeneous linear equations are obtained, and the basic solution system is the desired vector, then the known 3 vector Schmidt is normalized.

The eigenvalues and eigenvectors of the second square matrix I. Mathematical concepts

1. Characteristic values and eigenvectors:

Set A to n-order square matrix, if the number and n-dimensional 0-column vector x, so that the relational ax=λx set up, then the number λ is a matrix a eigenvalue, Non-zero vector x is called a correspondence and eigenvalues of the eigenvector.

2. Feature polynomial

3. Characteristic equation

Two. Principles, formulas and rules

1. Methods of finding eigenvalues and eigenvectors:

(1) (Practical to abstract matrix);

(2) (Practical to concrete matrix);

(3) (mainly for eigenvector).

2. Main formula

Set is a eigenvalue, X is a corresponding to the eigenvalues of the eigenvector, then there are

Note: Eigenvalues and eigenvectors refer to a reversible time.

3. Characteristic value and characteristic vector's property

Set is the n eigenvalues of A, then there are

1)

2)

3 The sufficient and necessary condition of a reversibility is that a does not have a 0 characteristic value.

4 The sufficient and necessary condition for a irreversibility is that a has 0 eigenvalues.

5) The eigenvalues corresponding to the different eigenvalues of a phalanx A are linearly independent. Three. Analysis of key points and difficulties

The focus of this section is to understand the concept of eigenvalues and eigenvectors, to find the eigenvalues and eigenvectors of a, and to master various methods of finding eigenvalues and eigenvectors.  The difficulty is the proof of the linear independent feature vectors corresponding to the different eigenvalues of the matrix A, and the methods of finding the eigenvalues and eigenvectors of a square matrix. Four. Typical examples

Example 1. Seeking Square matrix

Eigenvalues and eigenvectors.

Solution: A's feature polynomial is

，

So the eigenvalue of a is.

At that time, the solution equation (a-2e) x=0. By

，

A basic solution to the system

，

So it's all the eigenvector that corresponds to it.

When, the solution equation (a-e) x=0. By

A basic solution to the system

，

So it's all the eigenvector that corresponds to it.

Example 2. Finding matrices

Eigenvalues and eigenvectors.

Solution

So the eigenvalue of a is.

At that time, the solution equation (a+e) x=0. By

，

A basic solution to the system

，

So it's all the eigenvector that corresponds to it.

At that time, the solution equation (a-2e) x=0. By

，

A basic solution to the system

，

So all the eigenvectors correspond to the

。

The above example 1, example 2 all have the double characteristic value, in Example 1, the dual eigenvalue corresponds to two linear correlation eigenvectors, and in Example 2 the dual eigenvalues correspond to two linear eigenvectors, which is important for the diagonalization of the matrices to be studied below, hoping to arouse the attention of the students.

Example 3. Set a 3-order matrix A satisfied, and the rank of matrix A is 2, to find the eigenvalues of a.

Solution: Set is a eigenvalue, X is a of the corresponding eigenvector, then, at both ends to the right by X, get

That

That

Because, so

Have

The rank of a is 2, and the characteristic value of a is

Example 3 is an abstract matrix to find the eigenvalue problem, from the given known conditions, and then according to the constraints (example a rank equals 2) to determine the eigenvalues of a.

Section III similarity matrix I. Mathematical concepts

1. Similarity matrix:

Set A, B are n-order matrices, if there is a reversible matrix p, so that

B is called a similarity matrix, which is a~b.

2. Similar transformations

The operation of A is called the similarity transformation matrix for A. Two. Principles, formulas and rules

1. The similarity matrix has the same characteristic polynomial, which has the same characteristic value.

2. If a is similar to the diagonal matrix L, then the L-Main-line element is the N-eigenvalues of A.

3. The sufficient and necessary conditions for the N-order matrix A can be similar to that of the diagonal matrices are: A has n linearly independent eigenvectors.

4. If the n eigenvalues of N-order matrices A are different, a is similar to the diagonal matrix L.

5. The eigenvalues of a real symmetric matrix are real.

6. The characteristic vectors corresponding to the different eigenvalues of the real symmetric matrices are linearly independent.

7. Set is the real symmetric matrix a K-weight eigenvalue, the rank of the matrix, so that the K-weight eigenvalue has a K-linear independent eigenvector.

8. To set a n-order real symmetric matrix, there must be an orthogonal matrix p, where L is a diagonal matrix with the n eigenvalue of a as the diagonal element. Three. Analysis of key points and difficulties

The focus of this section is the condition that the general matrix can be diagonalization, the real symmetric matrix is transformed into diagonal matrices by orthogonal transformation, and a symmetric matrix is formed into diagonal matrices to lay the groundwork for the two-second form. The difficulty is the proof and derivation of the above theory, and how to use the orthogonal transformation matrix to transform symmetry into diagonal matrices. This kind of problem solving method has the very strong regularity, but the step is more, makes more complex, the schoolmates study is still more difficult. Four. Typical examples

Example 1. Set The Matrix

，

And A is similar to B, the value of the request.

Solution: Since A is similar to B, then

That

Again

GET, thus

The solution of such problems can be solved by the ①a of the equation, which is equal to the characteristic value: ②| A| is equal to the product of a characteristic value. If the above two equations are the same, they can be obtained by substituting the elements (that is, the eigenvalues of a) on the main diagonal of the diagonal matrix B.

Example 2. Set the Matrix

When k equals what value, there is a reversible matrix p, which makes. And the P and the corresponding diagonal matrices are obtained.

Solution: By

Have

Was

Obviously when k = 0 o'clock, the corresponding eigenvector is

，

Was

，

The corresponding eigenvector is

。

So when k=0, 0

The

。

The key analysis of this kind of problem is that a double eigenvalue should be given, and the two eigenvalues should correspond to two linearly independent eigenvectors. OK K, this is a very critical step.

Example 3. Set the Matrix

Find the orthogonal matrices P and L, so that.

Solution: By

The characteristic value of a is.

At that time, substituting equations, i.e.

，

When the solution is a characteristic vector

；

At that time, substituting equations, i.e.

，

The corresponding eigenvector of the solution is

Obviously with orthogonal, but linearly independent, can use the Schmidt standard orthogonal to the 22 orthogonal unit vector, which is more troublesome, if, obviously they are orthogonal, and is the solution of the above linear equation group, so only need to unit

，

Order, then p is an orthogonal matrix, and

。

From:http://dec3.jlu.edu.cn/webcourse/t000022/teach/index.htm