Find the operator's eigenvalues. Eigenvalues ​​(numbers) and eigenvectors. Examples of solutions

Diagonal matrices have the simplest structure. The question arises whether it is possible to find a basis in which the matrix of the linear operator would have a diagonal form. Such a basis exists.
Let us be given a linear space R n and a linear operator A acting in it; in this case, operator A takes R n into itself, that is, A:R n → R n .

Definition. A non-zero vector is called an eigenvector of the operator A if the operator A translates into a collinear vector, that is. The number λ is called the eigenvalue or eigenvalue of the operator A, corresponding to the eigenvector.
Let us note some properties of eigenvalues ​​and eigenvectors.
1. Any linear combination of eigenvectors operator A corresponding to the same eigenvalue λ is an eigenvector with the same eigenvalue.
2. Eigenvectors operator A with pairwise different eigenvalues ​​λ 1 , λ 2 , …, λ m are linearly independent.
3. If the eigenvalues ​​λ 1 =λ 2 = λ m = λ, then the eigenvalue λ corresponds to no more than m linearly independent eigenvectors.

So, if there are n linearly independent eigenvectors , corresponding to different eigenvalues ​​λ 1, λ 2, ..., λ n, then they are linearly independent, therefore, they can be taken as the basis of the space R n. Let us find the form of the matrix of the linear operator A in the basis of its eigenvectors, for which we will act with the operator A on the basis vectors: Then .
Thus, the matrix of the linear operator A in the basis of its eigenvectors has a diagonal form, and the eigenvalues ​​of the operator A are along the diagonal.
Is there another basis in which the matrix has a diagonal form? The answer to this question is given by the following theorem.

Theorem. The matrix of a linear operator A in the basis (i = 1..n) has a diagonal form if and only if all the vectors of the basis are eigenvectors of the operator A.

Rule for finding eigenvalues ​​and eigenvectors

. (*)

(1)
Where - linear operator matrix.

System (1) has a non-zero solution if its determinant D is equal to zero

Example 12. The linear operator A acts in R 3 according to the law, where x 1, x 2, .., x n are the coordinates of the vector in the basis , , . Find the eigenvalues ​​and eigenvectors of this operator.
Solution. We build the matrix of this operator:
.
We create a system for determining the coordinates of eigenvectors:

We compose a characteristic equation and solve it:

.
λ 1,2 = -1, λ 3 = 3.
Substituting λ = -1 into the system, we have:
or
Because , then there are two dependent variables and one free variable.
Let x 1 be a free unknown, then We solve this system in any way and find the general solution of this system: The fundamental system of solutions consists of one solution, since n - r = 3 - 2 = 1.
The set of eigenvectors corresponding to the eigenvalue λ = -1 has the form: , where x 1 is any number other than zero. Let's choose one vector from this set, for example, putting x 1 = 1: .
Reasoning similarly, we find the eigenvector corresponding to the eigenvalue λ = 3: .
In the space R 3, the basis consists of three linearly independent vectors, but we received only two linearly independent eigenvectors, from which the basis in R 3 cannot be composed. Consequently, we cannot reduce the matrix A of a linear operator to diagonal form.

Example 13. Given a matrix .
1. Prove that the vector is an eigenvector of matrix A. Find the eigenvalue corresponding to this eigenvector.
2. Find a basis in which matrix A has a diagonal form.
Solution.
1. If , then is an eigenvector

.
Vector (1, 8, -1) is an eigenvector. Eigenvalue λ = -1.
The matrix has a diagonal form in a basis consisting of eigenvectors. One of them is famous. Let's find the rest.
We look for eigenvectors from the system:

Characteristic equation: ;
(3 + λ)[-2(2-λ)(2+λ)+3] = 0; (3+λ)(λ 2 - 1) = 0
λ 1 = -3, λ 2 = 1, λ 3 = -1.
Let's find the eigenvector corresponding to the eigenvalue λ = -3:

The rank of the matrix of this system is two and equal to the number of unknowns, so this system has only a zero solution x 1 = x 3 = 0. x 2 here can be anything other than zero, for example, x 2 = 1. Thus, the vector (0 ,1,0) is an eigenvector corresponding to λ = -3. Let's check:
.
If λ = 1, then we obtain the system
The rank of the matrix is ​​two. We cross out the last equation.
Let x 3 be a free unknown. Then x 1 = -3x 3, 4x 2 = 10x 1 - 6x 3 = -30x 3 - 6x 3, x 2 = -9x 3.
Assuming x 3 = 1, we have (-3,-9,1) - an eigenvector corresponding to the eigenvalue λ = 1. Check:

.
Since the eigenvalues ​​are real and distinct, the vectors corresponding to them are linearly independent, so they can be taken as a basis in R 3 . Thus, in the basis , , matrix A has the form:
.
Not every matrix of a linear operator A:R n → R n can be reduced to diagonal form, since for some linear operators there may be less than n linear independent eigenvectors. However, if the matrix is ​​symmetric, then the root of the characteristic equation of multiplicity m corresponds to exactly m linearly independent vectors.

Definition. A symmetric matrix is ​​a square matrix in which the elements symmetric about the main diagonal are equal, that is, in which .
Notes. 1. All eigenvalues ​​of a symmetric matrix are real.
2. The eigenvectors of a symmetric matrix corresponding to pairwise different eigenvalues ​​are orthogonal.
As one of the many applications of the studied apparatus, we consider the problem of determining the type of a second-order curve.

Eigenvalues ​​(numbers) and eigenvectors.
Examples of solutions

Be yourself

From both equations it follows that .

Let's put it then: .

As a result: – second eigenvector.

Let us repeat the important points of the decision:

– the resulting system certainly has a general solution (the equations are linearly dependent);

– we select the “y” in such a way that it is integer and the first “x” coordinate is integer, positive and as small as possible.

– we check that the particular solution satisfies each equation of the system.

Answer .

There were quite enough intermediate “checkpoints”, so checking equality is, in principle, unnecessary.

In various sources of information, the coordinates of eigenvectors are often written not in columns, but in rows, for example: (and, to be honest, I myself am used to writing them down in lines). This option is acceptable, but in light of the topic linear transformations technically more convenient to use column vectors.

Perhaps the solution seemed very long to you, but this is only because I commented on the first example in great detail.

Example 2

Matrices

Let's train on our own! An approximate example of a final task at the end of the lesson.

Sometimes you need to complete an additional task, namely:

write the canonical matrix decomposition

What it is?

If the eigenvectors of the matrix form basis, then it can be represented as:

Where is a matrix composed of coordinates of eigenvectors, – diagonal matrix with corresponding eigenvalues.

This matrix decomposition is called canonical or diagonal.

Let's look at the matrix of the first example. Its eigenvectors linearly independent(non-collinear) and form a basis. Let's create a matrix of their coordinates:

On main diagonal matrices in the appropriate order the eigenvalues ​​are located, and the remaining elements are equal to zero:
– I once again emphasize the importance of order: “two” corresponds to the 1st vector and is therefore located in the 1st column, “three” – to the 2nd vector.

Using the usual algorithm for finding inverse matrix or Gauss-Jordan method we find . No, that's not a typo! - before you is a rare event, like a solar eclipse, when the reverse coincided with the original matrix.

It remains to write down the canonical decomposition of the matrix:

The system can be solved using elementary transformations, and in the following examples we will resort to this method. But here the “school” method works much faster. From the 3rd equation we express: – substitute into the second equation:

Since the first coordinate is zero, we obtain a system, from each equation of which it follows that .

And again pay attention to the mandatory presence of a linear relationship. If only a trivial solution is obtained , then either the eigenvalue was found incorrectly, or the system was compiled/solved with an error.

Compact coordinates gives the value

Eigenvector:

And once again, we check that the solution found satisfies every equation of the system. In subsequent paragraphs and in subsequent tasks, I recommend taking this wish as a mandatory rule.

2) For the eigenvalue, using the same principle, we obtain the following system:

From the 2nd equation of the system we express: – substitute into the third equation:

Since the “zeta” coordinate is equal to zero, we obtain a system from each equation of which a linear dependence follows.

Let

Checking that the solution satisfies every equation of the system.

Thus, the eigenvector is: .

3) And finally, the system corresponds to the eigenvalue:

The second equation looks the simplest, so let’s express it and substitute it into the 1st and 3rd equations:

Everything is fine - a linear relationship has emerged, which we substitute into the expression:

As a result, “x” and “y” were expressed through “z”: . In practice, it is not necessary to achieve precisely such relationships; in some cases it is more convenient to express both through or and through . Or even “train” - for example, “X” through “I”, and “I” through “Z”

Let's put it then:

We check that the solution found satisfies each equation of the system and writes the third eigenvector

Answer: eigenvectors:

Geometrically, these vectors define three different spatial directions ("There and back again"), according to which linear transformation transforms non-zero vectors (eigenvectors) into collinear vectors.

If the condition required finding the canonical decomposition, then this is possible here, because different eigenvalues ​​correspond to different linearly independent eigenvectors. Making a matrix from their coordinates, a diagonal matrix from relevant eigenvalues ​​and find inverse matrix .

If, by condition, you need to write linear transformation matrix in the basis of eigenvectors, then we give the answer in the form . There is a difference, and the difference is significant! Because this matrix is ​​the “de” matrix.

A problem with simpler calculations for you to solve on your own:

Example 5

Find eigenvectors of a linear transformation given by a matrix

When finding your own numbers, try not to go all the way to a 3rd degree polynomial. In addition, your system solutions may differ from my solutions - there is no certainty here; and the vectors you find may differ from the sample vectors up to the proportionality of their respective coordinates. For example, and. It is more aesthetically pleasing to present the answer in the form, but it’s okay if you stop at the second option. However, there are reasonable limits to everything; the version no longer looks very good.

An approximate final sample of the assignment at the end of the lesson.

How to solve the problem in the case of multiple eigenvalues?

The general algorithm remains the same, but it has its own characteristics, and it is advisable to keep some parts of the solution in a more strict academic style:

Example 6

Find eigenvalues ​​and eigenvectors

Solution

Of course, let’s capitalize the fabulous first column:

And, after factoring the quadratic trinomial:

As a result, eigenvalues ​​are obtained, two of which are multiples.

Let's find the eigenvectors:

1) Let’s deal with a lone soldier according to a “simplified” scheme:

From the last two equations, the equality is clearly visible, which, obviously, should be substituted into the 1st equation of the system:

You won't find a better combination:
Eigenvector:

2-3) Now we remove a couple of sentries. In this case it may turn out either two or one eigenvector. Regardless of the multiplicity of the roots, we substitute the value into the determinant which brings us the next homogeneous system of linear equations:

Eigenvectors are exactly vectors
fundamental system of solutions

Actually, throughout the entire lesson we did nothing but find the vectors of the fundamental system. It’s just that for the time being this term was not particularly required. By the way, those clever students who missed the topic in camouflage suits homogeneous equations, will be forced to smoke it now.

The only action was to remove the extra lines. The result is a one-by-three matrix with a formal “step” in the middle.
– basic variable, – free variables. There are two free variables, therefore there are also two vectors of the fundamental system.

Let's express the basic variable in terms of free variables: . The zero multiplier in front of the “X” allows it to take on absolutely any values ​​(which is clearly visible from the system of equations).

In the context of this problem, it is more convenient to write the general solution not in a row, but in a column:

The pair corresponds to an eigenvector:
The pair corresponds to an eigenvector:

Note : sophisticated readers can select these vectors orally - simply by analyzing the system , but some knowledge is needed here: there are three variables, system matrix rank- one, which means fundamental decision system consists of 3 – 1 = 2 vectors. However, the found vectors are clearly visible even without this knowledge, purely on an intuitive level. In this case, the third vector will be written even more “beautifully”: . However, I warn you that in another example, a simple selection may not be possible, which is why the clause is intended for experienced people. In addition, why not take, say, as the third vector? After all, its coordinates also satisfy each equation of the system, and the vectors linearly independent. This option, in principle, is suitable, but “crooked”, since the “other” vector is a linear combination of vectors of the fundamental system.

Answer: eigenvalues: , eigenvectors:

A similar example for an independent solution:

Example 7

Find eigenvalues ​​and eigenvectors

An approximate sample of the final design at the end of the lesson.

It should be noted that in both the 6th and 7th examples a triple of linearly independent eigenvectors is obtained, and therefore the original matrix is ​​representable in the canonical decomposition. But such raspberries do not happen in all cases:

Example 8

Solution: Let’s create and solve the characteristic equation:

Let's expand the determinant in the first column:

We carry out further simplifications according to the considered method, avoiding the third-degree polynomial:

– eigenvalues.

Let's find the eigenvectors:

1) There are no difficulties with the root:

Don’t be surprised, in addition to the kit, there are also variables in use - there is no difference here.

From the 3rd equation we express it and substitute it into the 1st and 2nd equations:

From both equations it follows:

Let then:

2-3) For multiple values ​​we get the system .

Let's write down the matrix of the system and, using elementary transformations, bring it to a stepwise form:

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An eigenvector of a square matrix is ​​one that, when multiplied by a given matrix, results in a collinear vector. In simple words, when a matrix is ​​multiplied by an eigenvector, the latter remains the same, but multiplied by a certain number.

Definition

An eigenvector is a non-zero vector V, which, when multiplied by a square matrix M, becomes itself increased by some number λ. In algebraic notation it looks like:

M × V = λ × V,

where λ is the eigenvalue of the matrix M.

Let's look at a numerical example. For ease of recording, numbers in the matrix will be separated by a semicolon. Let us have a matrix:

• M = 0; 4;
• 6; 10.

Let's multiply it by a column vector:

• V = -2;

When we multiply a matrix by a column vector, we also get a column vector. In strict mathematical language, the formula for multiplying a 2 × 2 matrix by a column vector will look like this:

• M × V = M11 × V11 + M12 × V21;
• M21 × V11 + M22 × V21.

M11 means the element of matrix M located in the first row and first column, and M22 means the element located in the second row and second column. For our matrix, these elements are equal to M11 = 0, M12 = 4, M21 = 6, M22 10. For a column vector, these values ​​are equal to V11 = –2, V21 = 1. According to this formula, we get the following result of the product of a square matrix by a vector:

• M × V = 0 × (-2) + (4) × (1) = 4;
• 6 × (-2) + 10 × (1) = -2.

For convenience, let's write the column vector into a row. So, we multiplied the square matrix by the vector (-2; 1), resulting in the vector (4; -2). Obviously, this is the same vector multiplied by λ = -2. Lambda in this case denotes the eigenvalue of the matrix.

An eigenvector of a matrix is ​​a collinear vector, that is, an object that does not change its position in space when multiplied by a matrix. The concept of collinearity in vector algebra is similar to the term of parallelism in geometry. In a geometric interpretation, collinear vectors are parallel directed segments of different lengths. Since the time of Euclid, we know that one line has an infinite number of lines parallel to it, so it is logical to assume that each matrix has an infinite number of eigenvectors.

From the previous example it is clear that eigenvectors can be (-8; 4), and (16; -8), and (32, -16). These are all collinear vectors corresponding to the eigenvalue λ = -2. When multiplying the original matrix by these vectors, we will still end up with a vector that differs from the original by 2 times. That is why, when solving problems of finding an eigenvector, it is necessary to find only linearly independent vector objects. Most often, for an n × n matrix, there are an n number of eigenvectors. Our calculator is designed for the analysis of second-order square matrices, so almost always the result will find two eigenvectors, except for cases when they coincide.

In the example above, we knew the eigenvector of the original matrix in advance and clearly determined the lambda number. However, in practice, everything happens the other way around: the eigenvalues ​​are found first and only then the eigenvectors.

Solution algorithm

Let's look at the original matrix M again and try to find both of its eigenvectors. So the matrix looks like:

• M = 0; 4;
• 6; 10.

First we need to determine the eigenvalue λ, which requires calculating the determinant of the following matrix:

• (0 − λ); 4;
• 6; (10 − λ).

This matrix is ​​obtained by subtracting the unknown λ from the elements on the main diagonal. The determinant is determined using the standard formula:

• detA = M11 × M21 − M12 × M22
• detA = (0 − λ) × (10 − λ) − 24

Since our vector must be non-zero, we accept the resulting equation as linearly dependent and equate our determinant detA to zero.

(0 − λ) × (10 − λ) − 24 = 0

Let's open the brackets and get the characteristic equation of the matrix:

λ 2 − 10λ − 24 = 0

This is a standard quadratic equation that needs to be solved using a discriminant.

D = b 2 − 4ac = (-10) × 2 − 4 × (-1) × 24 = 100 + 96 = 196

The root of the discriminant is sqrt(D) = 14, therefore λ1 = -2, λ2 = 12. Now for each lambda value we need to find the eigenvector. Let us express the system coefficients for λ = -2.

• M − λ × E = 2; 4;
• 6; 12.

In this formula, E is the identity matrix. Based on the resulting matrix, we create a system of linear equations:

2x + 4y = 6x + 12y,

where x and y are the eigenvector elements.

Let's collect all the X's on the left and all the Y's on the right. Obviously - 4x = 8y. Divide the expression by - 4 and get x = –2y. Now we can determine the first eigenvector of the matrix, taking any values ​​of the unknowns (remember the infinity of linearly dependent eigenvectors). Let's take y = 1, then x = –2. Therefore, the first eigenvector looks like V1 = (–2; 1). Return to the beginning of the article. It was this vector object that we multiplied the matrix by to demonstrate the concept of an eigenvector.

Now let's find the eigenvector for λ = 12.

• M - λ × E = -12; 4
• 6; -2.

Let's create the same system of linear equations;

• -12x + 4y = 6x − 2y
• -18x = -6y
• 3x = y.

Now we take x = 1, therefore y = 3. Thus, the second eigenvector looks like V2 = (1; 3). When multiplying the original matrix by a given vector, the result will always be the same vector multiplied by 12. This is where the solution algorithm ends. Now you know how to manually determine the eigenvector of a matrix.

• determinant;
• trace, that is, the sum of the elements on the main diagonal;
• rank, that is, the maximum number of linearly independent rows/columns.

The program operates according to the above algorithm, shortening the solution process as much as possible. It is important to point out that in the program lambda is designated by the letter “c”. Let's look at a numerical example.

Example of how the program works

Let's try to determine the eigenvectors for the following matrix:

• M = 5; 13;
• 4; 14.

Let's enter these values ​​into the cells of the calculator and get the answer in the following form:

• Matrix rank: 2;
• Matrix determinant: 18;
• Matrix trace: 19;
• Calculation of the eigenvector: c 2 − 19.00c + 18.00 (characteristic equation);
• Eigenvector calculation: 18 (first lambda value);
• Eigenvector calculation: 1 (second lambda value);
• System of equations for vector 1: -13x1 + 13y1 = 4x1 − 4y1;
• System of equations for vector 2: 4x1 + 13y1 = 4x1 + 13y1;
• Eigenvector 1: (1; 1);
• Eigenvector 2: (-3.25; 1).

Thus, we obtained two linearly independent eigenvectors.

Conclusion

Linear algebra and analytical geometry are standard subjects for any freshman engineering major. The large number of vectors and matrices is terrifying, and it is easy to make mistakes in such cumbersome calculations. Our program will allow students to check their calculations or automatically solve the problem of finding an eigenvector. There are other linear algebra calculators in our catalog; use them in your studies or work.

"The first part sets out the provisions that are minimally necessary for understanding chemometrics, and the second part contains the facts that you need to know for a deeper understanding of the methods of multivariate analysis. The presentation is illustrated with examples made in the Excel workbook Matrix.xls, which accompanies this document.

Links to examples are placed in the text as Excel objects. These examples are of an abstract nature; they are in no way tied to the problems of analytical chemistry. Real-life examples of the use of matrix algebra in chemometrics are discussed in other texts covering a variety of chemometric applications.

Most measurements made in analytical chemistry are not direct, but indirect. This means that in the experiment, instead of the value of the desired analyte C (concentration), another value is obtained x(signal), related but not equal to C, i.e. x(C) ≠ C. As a rule, the type of dependence x(C) is unknown, but fortunately in analytical chemistry most measurements are proportional. This means that with increasing concentration of C in a times, signal X will increase by the same amount, i.e. x(a C) = a x(C). In addition, the signals are also additive, so the signal from a sample in which two substances with concentrations C 1 and C 2 are present will be equal to the sum of the signals from each component, i.e. x(C 1 + C 2) = x(C 1)+ x(C 2). Proportionality and additivity together give linearity. Many examples can be given to illustrate the principle of linearity, but it is enough to mention the two most striking examples - chromatography and spectroscopy. The second feature inherent in an experiment in analytical chemistry is multichannel. Modern analytical equipment simultaneously measures signals for many channels. For example, the intensity of light transmission is measured for several wavelengths at once, i.e. range. Therefore, in the experiment we deal with many signals x 1 , x 2 ,...., x n, characterizing the set of concentrations C 1 , C 2 , ..., C m of substances present in the system under study.

Rice. 1 Spectra

So, an analytical experiment is characterized by linearity and multidimensionality. Therefore, it is convenient to consider experimental data as vectors and matrices and manipulate them using the apparatus of matrix algebra. The fruitfulness of this approach is illustrated by the example shown in, which presents three spectra taken at 200 wavelengths from 4000 to 4796 cm −1. First ( x 1) and second ( x 2) the spectra were obtained for standard samples in which the concentrations of two substances A and B are known: in the first sample [A] = 0.5, [B] = 0.1, and in the second sample [A] = 0.2, [B] = 0.6. What can be said about a new, unknown sample, the spectrum of which is indicated x 3 ?

Let us consider three experimental spectra x 1 , x 2 and x 3 as three vectors of dimension 200. Using linear algebra, one can easily show that x 3 = 0.1 x 1 +0.3 x 2, so the third sample obviously contains only substances A and B in concentrations [A] = 0.5×0.1 + 0.2×0.3 = 0.11 and [B] = 0.1×0.1 + 0.6×0.3 = 0.19.

1. Basic information

1.1 Matrices

Matrix called a rectangular table of numbers, for example

Rice. 2 Matrix

Matrices are denoted by capital bold letters ( A), and their elements - by corresponding lowercase letters with indices, i.e. a ij. The first index numbers the rows, and the second - the columns. In chemometrics, it is customary to denote the maximum value of an index with the same letter as the index itself, but in capital letters. Therefore the matrix A can also be written as ( a ij , i = 1,..., I; j = 1,..., J). For the example matrix I = 4, J= 3 and a 23 = −7.5.

Pair of numbers I And J is called the dimension of the matrix and is denoted as I× J. An example of a matrix in chemometrics is a set of spectra obtained for I samples for J wavelengths.

1.2. The simplest operations with matrices

Matrices can be multiply by numbers. In this case, each element is multiplied by this number. For example -

Rice. 3 Multiplying a matrix by a number

Two matrices of the same dimension can be element by element fold And subtract. For example,

Rice. 4 Matrix addition

As a result of multiplication by a number and addition, a matrix of the same dimension is obtained.

A zero matrix is ​​a matrix consisting of zeros. It is designated O. It's obvious that A+O = A, A−A = O and 0 A = O.

The matrix can be transpose. During this operation, the matrix is ​​flipped, i.e. rows and columns are swapped. Transposition is indicated by a prime, A" or index A t. Thus, if A = {a ij , i = 1,..., I; j = 1,...,J), That A t = ( a ji , j = 1,...,J; i = 1,..., I). For example

Rice. 5 Matrix transposition

It's obvious that ( A t) t = A, (A+B) t =A t+ B t.

1.3. Matrix multiplication

Matrices can be multiply, but only if they have the appropriate dimensions. Why this is so will be clear from the definition. Matrix product A, dimension I× K, and matrices B, dimension K× J, is called a matrix C, dimension I× J, whose elements are numbers

Thus for the product AB it is necessary that the number of columns in the left matrix A was equal to the number of rows in the right matrix B. An example of a matrix product -

Fig.6 Product of matrices

The rule for matrix multiplication can be formulated as follows. To find a matrix element C, standing at the intersection i-th line and j th column ( c ij) must be multiplied element by element i-th row of the first matrix A on j th column of the second matrix B and add up all the results. So in the example shown, an element from the third row and the second column is obtained as the sum of the element-wise products of the third row A and second column B

Fig.7 Element of the product of matrices

The product of matrices depends on the order, i.e. AB ≠ B.A., at least for dimensional reasons. They say that it is non-commutative. However, the product of matrices is associative. It means that ABC = (AB)C = A(B.C.). In addition, it is also distributive, i.e. A(B+C) = AB+A.C.. It's obvious that A.O. = O.

1.4. Square matrices

If the number of matrix columns is equal to the number of its rows ( I = J=N), then such a matrix is ​​called square. In this section we will consider only such matrices. Among these matrices, matrices with special properties can be distinguished.

Single matrix (denoted I, and sometimes E) is a matrix in which all elements are equal to zero, with the exception of diagonal ones, which are equal to 1, i.e.

Obviously A.I. = I.A. = A.

The matrix is ​​called diagonal, if all its elements except diagonal ones ( a ii) are equal to zero. For example

Rice. 8 Diagonal matrix

Matrix A called the top triangular, if all its elements lying below the diagonal are equal to zero, i.e. a ij= 0, at i>j. For example

Rice. 9 Upper triangular matrix

The lower triangular matrix is ​​defined similarly.

Matrix A called symmetrical, If A t = A. In other words a ij = a ji. For example

Rice. 10 Symmetric matrix

Matrix A called orthogonal, If

A t A = A.A. t = I.

The matrix is ​​called normal If

1.5. Trace and determinant

Next square matrix A(denoted by Tr( A) or Sp( A)) is the sum of its diagonal elements,

For example,

Rice. 11 Matrix trace

It's obvious that

Sp(α A) = α Sp( A) And

Sp( A+B) = Sp( A)+ Sp( B).

It can be shown that

Sp( A) = Sp( A t), Sp( I) = N,

and also that

Sp( AB) = Sp( B.A.).

Another important characteristic of a square matrix is ​​its determinant(denoted det( A)). Determining the determinant in the general case is quite difficult, so we will start with the simplest option - the matrix A dimension (2×2). Then

For a (3×3) matrix the determinant will be equal to

In the case of the matrix ( N× N) the determinant is calculated as the sum 1·2·3· ... · N= N! terms, each of which is equal

Indexes k 1 , k 2 ,..., k N are defined as all possible ordered permutations r numbers in the set (1, 2, ..., N). Calculating the determinant of a matrix is ​​a complex procedure, which in practice is carried out using special programs. For example,

Rice. 12 Matrix determinant

Let us note only the obvious properties:

det( I) = 1, det( A) = det( A t),

det( AB) = det( A)det( B).

1.6. Vectors

If the matrix consists of only one column ( J= 1), then such an object is called vector. More precisely, a column vector. For example

One can also consider matrices consisting of one row, for example

This object is also a vector, but row vector. When analyzing data, it is important to understand which vectors we are dealing with - columns or rows. So the spectrum taken for one sample can be considered as a row vector. Then the set of spectral intensities at a certain wavelength for all samples should be treated as a column vector.

The dimension of a vector is the number of its elements.

It is clear that any column vector can be turned into a row vector by transposition, i.e.

In cases where the shape of the vector is not specifically specified, but is simply said to be a vector, then they mean a column vector. We will also adhere to this rule. A vector is denoted by a lowercase, upright, bold letter. A zero vector is a vector all of whose elements are zero. It is designated 0 .

1.7. The simplest operations with vectors

Vectors can be added and multiplied by numbers in the same way as matrices. For example,

Rice. 13 Operations with vectors

Two vectors x And y are called colinear, if there is a number α such that

1.8. Products of vectors

Two vectors of the same dimension N can be multiplied. Let there be two vectors x = (x 1 , x 2 ,...,x N)t and y = (y 1 , y 2 ,...,y N) t . Guided by the row-by-column multiplication rule, we can compose two products from them: x t y And xy t. First work

called scalar or internal. Its result is a number. It is also denoted by ( x,y)= x t y. For example,

Rice. 14 Inner (scalar) product

Second piece

called external. Its result is a matrix of dimension ( N× N). For example,

Rice. 15 External work

Vectors whose scalar product is zero are called orthogonal.

1.9. Vector norm

The scalar product of a vector with itself is called a scalar square. This value

defines a square length vector x. To indicate length (also called the norm vector) the notation is used

For example,

Rice. 16 Vector norm

Unit length vector (|| x|| = 1) is called normalized. Non-zero vector ( x ≠ 0 ) can be normalized by dividing it by length, i.e. x = ||x|| (x/||x||) = ||x|| e. Here e = x/||x|| - normalized vector.

Vectors are called orthonormal if they are all normalized and pairwise orthogonal.

1.10. Angle between vectors

The scalar product determines and cornerφ between two vectors x And y

If the vectors are orthogonal, then cosφ = 0 and φ = π/2, and if they are colinear, then cosφ = 1 and φ = 0.

1.11. Vector representation of a matrix

Each matrix A size I× J can be represented as a set of vectors

Here every vector a j is j th column, and the row vector b i is i th row of the matrix A

1.12. Linearly dependent vectors

Vectors of the same dimension ( N) can be added and multiplied by a number, just like matrices. The result will be a vector of the same dimension. Let there be several vectors of the same dimension x 1 , x 2 ,...,x K and the same number of numbers α α 1 , α 2 ,...,α K. Vector

y= α 1 x 1 + α 2 x 2 +...+ α K x K

called linear combination vectors x k .

If there are such non-zero numbers α k ≠ 0, k = 1,..., K, What y = 0 , then such a set of vectors x k called linearly dependent. Otherwise, the vectors are said to be linearly independent. For example, vectors x 1 = (2, 2)t and x 2 = (−1, −1) t are linearly dependent, because x 1 +2x 2 = 0

1.13. Matrix rank

Consider a set of K vectors x 1 , x 2 ,...,x K dimensions N. The rank of this system of vectors is the maximum number of linearly independent vectors. For example in the set

there are only two linearly independent vectors, for example x 1 and x 2, so its rank is 2.

Obviously, if there are more vectors in a set than their dimension ( K>N), then they are necessarily linearly dependent.

Matrix rank(denoted by rank( A)) is the rank of the system of vectors of which it consists. Although any matrix can be represented in two ways (column or row vectors), this does not affect the rank value, because

1.14. inverse matrix

Square matrix A is called non-degenerate if it has a unique reverse matrix A-1, determined by the conditions

A.A. −1 = A −1 A = I.

The inverse matrix does not exist for all matrices. A necessary and sufficient condition for non-degeneracy is

det( A) ≠ 0 or rank( A) = N.

Matrix inversion is a complex procedure for which there are special programs. For example,

Rice. 17 Matrix inversion

Let us present the formulas for the simplest case - a 2×2 matrix

If matrices A And B are non-degenerate, then

(AB) −1 = B −1 A −1 .

1.15. Pseudoinverse matrix

If matrix A is singular and the inverse matrix does not exist, then in some cases you can use pseudoinverse matrix, which is defined as such a matrix A+ that

A.A. + A = A.

The pseudoinverse matrix is ​​not the only one and its form depends on the method of construction. For example, for a rectangular matrix you can use the Moore-Penrose method.

If the number of columns is less than the number of rows, then

A + =(A t A) −1 A t

For example,

Rice. 17a Pseudo-inversion of a matrix

If the number of columns is greater than the number of rows, then

A + =A t ( A.A. t) −1

1.16. Multiplying a vector by a matrix

Vector x can be multiplied by a matrix A suitable size. In this case, the column vector is multiplied on the right Ax, and the vector row is on the left x t A. If the vector dimension J, and the matrix dimension I× J then the result will be a vector of dimension I. For example,

Rice. 18 Multiplying a vector by a matrix

If matrix A- square ( I× I), then the vector y = Ax has the same dimension as x. It's obvious that

A(α 1 x 1 + α 2 x 2) = α 1 Ax 1 + α 2 Ax 2 .

Therefore, matrices can be considered as linear transformations of vectors. In particular Ix = x, Ox = 0 .

2. Additional information

2.1. Systems of linear equations

Let A- matrix size I× J, A b- dimension vector J. Consider the equation

Ax = b

relative to the vector x, dimensions I. Essentially, it is a system of I linear equations with J unknown x 1 ,...,x J. A solution exists if and only if

rank( A) = rank( B) = R,

Where B is an extended matrix of dimensions I×( J+1), consisting of a matrix A, supplemented by a column b, B = (A b). Otherwise, the equations are inconsistent.

If R = I = J, then the solution is unique

x = A −1 b.

If R < I, then there are many different solutions that can be expressed through a linear combination J−R vectors. System of homogeneous equations Ax = 0 with square matrix A (N× N) has a nontrivial solution ( x ≠ 0 ) if and only if det( A) = 0. If R= rank( A)<N, then there are N−R linearly independent solutions.

2.2. Bilinear and quadratic forms

If A is a square matrix, and x And y- vector of the corresponding dimension, then the scalar product of the form x t Ay called bilinear form defined by matrix A. At x = y expression x t Ax called quadratic form.

2.3. Positive definite matrices

Square matrix A called positive definite, if for any nonzero vector x ≠ 0 ,

x t Ax > 0.

Similarly defined negative (x t Ax < 0), non-negative (x t Ax≥ 0) and negative (x t Ax≤ 0) certain matrices.

2.4. Cholesky decomposition

If the symmetric matrix A is positive definite, then there is a unique triangular matrix U with positive elements, for which

A = U t U.

For example,

Rice. 19 Cholesky decomposition

2.5. Polar decomposition

Let A is a non-singular square matrix of dimension N× N. Then there is a unique polar performance

A = S.R.

Where S is a non-negative symmetric matrix, and R is an orthogonal matrix. Matrices S And R can be defined explicitly:

S 2 = A.A. t or S = (A.A. t) ½ and R = S −1 A = (A.A. t) −½ A.

For example,

Rice. 20 Polar decomposition

If matrix A is degenerate, then the decomposition is not unique - namely: S still alone, but R maybe a lot. Polar decomposition represents the matrix A as a combination of compression/extension S and turn R.

2.6. Eigenvectors and eigenvalues

Let A is a square matrix. Vector v called eigenvector matrices A, If

Av = λ v,

where the number λ is called eigenvalue matrices A. Thus, the transformation that the matrix performs A above the vector v, comes down to simple stretching or compression with a coefficient λ. The eigenvector is determined up to multiplication by a constant α ≠ 0, i.e. If v is an eigenvector, then α v- also an eigenvector.

2.7. Eigenvalues

At the matrix A, dimension ( N× N) cannot be more than N eigenvalues. They satisfy characteristic equation

det( A − λ I) = 0,

which is an algebraic equation N-th order. In particular, for a 2×2 matrix the characteristic equation has the form

For example,

Rice. 21 Eigenvalues

Set of eigenvalues ​​λ 1 ,..., λ N matrices A called spectrum A.

The spectrum has various properties. In particular

det( A) = λ 1 ×...×λ N,Sp( A) = λ 1 +...+λ N.

The eigenvalues ​​of an arbitrary matrix can be complex numbers, but if the matrix is ​​symmetric ( A t = A), then its eigenvalues ​​are real.

2.8. Eigenvectors

At the matrix A, dimension ( N× N) cannot be more than N eigenvectors, each of which corresponds to its own eigenvalue. To determine the eigenvector v n need to solve a system of homogeneous equations

(A − λ n I)v n = 0 .

It has a non-trivial solution, since det( A −λ n I) = 0.

For example,

Rice. 22 Eigenvectors

The eigenvectors of a symmetric matrix are orthogonal.