To work with the concept of matrix rank, we will need information from the topic "Algebraic complements and minors. Types of minors and algebraic complements." First of all, this concerns the term “matrix minor”, since we will determine the rank of the matrix precisely through the minors.
Matrix rank is the maximum order of its minors, among which there is at least one that is not equal to zero.
Equivalent matrices- matrices whose ranks are equal to each other.
Let us explain in more detail. Suppose that among the second-order minors there is at least one that is different from zero. And all minors whose order is higher than two are equal to zero. Conclusion: the rank of the matrix is 2. Or, for example, among the minors of the tenth order there is at least one that is not equal to zero. And all minors whose order is higher than 10 are equal to zero. Conclusion: the rank of the matrix is 10.
The rank of the matrix $A$ is denoted as follows: $\rang A$ or $r(A)$. The rank of the zero matrix $O$ is assumed to be zero, $\rang O=0$. Let me remind you that to form a matrix minor you need to cross out rows and columns, but it is impossible to cross out more rows and columns than the matrix itself contains. For example, if the matrix $F$ has size $5\times 4$ (i.e. contains 5 rows and 4 columns), then the maximum order of its minors is four. It will no longer be possible to form minors of the fifth order, since they will require 5 columns (and we have only 4). This means that the rank of the matrix $F$ cannot be more than four, i.e. $\rang F≤4$.
In more general form, the above means that if a matrix contains $m$ rows and $n$ columns, then its rank cannot exceed the smallest of $m$ and $n$, i.e. $\rang A≤\min(m,n)$.
In principle, from the very definition of rank follows the method for finding it. The process of finding the rank of a matrix, by definition, can be schematically represented as follows:
Let me explain this diagram in more detail. Let's start reasoning from the very beginning, i.e. from the first order minors of some matrix $A$.
- If all first-order minors (i.e., elements of the matrix $A$) are equal to zero, then $\rang A=0$. If among the first-order minors there is at least one that is not equal to zero, then $\rang A≥ 1$. Let's move on to checking second-order minors.
- If all second-order minors are equal to zero, then $\rang A=1$. If among the second-order minors there is at least one that is not equal to zero, then $\rang A≥ 2$. Let's move on to checking third-order minors.
- If all third-order minors are equal to zero, then $\rang A=2$. If among the third-order minors there is at least one that is not equal to zero, then $\rang A≥ 3$. Let's move on to checking fourth-order minors.
- If all fourth-order minors are equal to zero, then $\rang A=3$. If among the fourth-order minors there is at least one that is not equal to zero, then $\rang A≥ 4$. We move on to checking fifth-order minors and so on.
What awaits us at the end of this procedure? It is possible that among the kth order minors there will be at least one that is different from zero, and all (k+1) order minors will be equal to zero. This means that k is the maximum order of minors, among which there is at least one that is not equal to zero, i.e. the rank will be equal to k. There may be a different situation: among the kth order minors there will be at least one that is not equal to zero, but it will no longer be possible to form (k+1) order minors. In this case, the rank of the matrix is also equal to k. In short, the order of the last composed non-zero minor will be equal to the rank of the matrix.
Let's move on to examples in which the process of finding the rank of a matrix, by definition, will be clearly illustrated. Let me emphasize once again that in the examples of this topic we will begin to find the rank of matrices using only the definition of rank. Other methods (calculating the rank of a matrix using the method of bordering minors, calculating the rank of a matrix using the method of elementary transformations) are discussed in the following topics.
By the way, it is not at all necessary to start the procedure for finding the rank with minors of the smallest order, as was done in examples No. 1 and No. 2. You can immediately move on to minors of higher orders (see example No. 3).
Example No. 1
Find the rank of the matrix $A=\left(\begin(array)(ccccc) 5 & 0 & -3 & 0 & 2 \\ 7 & 0 & -4 & 0 & 3 \\ 2 & 0 & -1 & 0 & 1 \end(array) \right)$.
This matrix has size $3\times 5$, i.e. contains three rows and five columns. Of the numbers 3 and 5, the minimum is 3, therefore the rank of the matrix $A$ is no more than 3, i.e. $\rang A≤ 3$. And this inequality is obvious, since we will no longer be able to form fourth-order minors - they require 4 rows, and we have only 3. Let’s move on directly to the process of finding the rank of a given matrix.
Among the first order minors (i.e. among the elements of the matrix $A$) there are non-zero ones. For example, 5, -3, 2, 7. In general, we are not interested in the total number of non-zero elements. There is at least one non-zero element - and that's enough. Since among the first-order minors there is at least one non-zero, we conclude that $\rang A≥ 1$ and proceed to checking the second-order minors.
Let's start exploring second order minors. For example, at the intersection of rows No. 1, No. 2 and columns No. 1, No. 4 there are elements of the following minor: $\left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right| $. For this determinant, all elements of the second column are equal to zero, therefore the determinant itself is equal to zero, i.e. $\left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right|=0$ (see property No. 3 in the topic of properties of determinants). Or you can simply calculate this determinant using formula No. 1 from the section on calculating second- and third-order determinants:
$$ \left|\begin(array)(cc) 5 & 0 \\ 7 & 0 \end(array) \right|=5\cdot 0-0\cdot 7=0. $$
The first second-order minor we tested turned out to be equal to zero. What does this mean? About the need to further check second-order minors. Either they will all turn out to be zero (and then the rank will be equal to 1), or among them there will be at least one minor that is different from zero. Let's try to make a better choice by writing a second-order minor, the elements of which are located at the intersection of rows No. 1, No. 2 and columns No. 1 and No. 5: $\left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array) \right|$. Let's find the value of this second-order minor:
$$ \left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array) \right|=5\cdot 3-2\cdot 7=1. $$
This minor is not equal to zero. Conclusion: among the second-order minors there is at least one non-zero. Therefore $\rang A≥ 2$. We need to move on to studying third-order minors.
If we choose column No. 2 or column No. 4 to form third-order minors, then such minors will be equal to zero (since they will contain a zero column). It remains to check only one third-order minor, the elements of which are located at the intersection of columns No. 1, No. 3, No. 5 and rows No. 1, No. 2, No. 3. Let's write down this minor and find its value:
$$ \left|\begin(array)(ccc) 5 & -3 & 2 \\ 7 & -4 & 3 \\ 2 & -1 & 1 \end(array) \right|=-20-18-14 +16+21+15=0. $$
So, all third order minors are equal to zero. The last non-zero minor we compiled was of second order. Conclusion: the maximum order of minors, among which there is at least one non-zero, is 2. Therefore, $\rang A=2$.
Answer: $\rang A=2$.
Example No. 2
Find the rank of the matrix $A=\left(\begin(array) (cccc) -1 & 3 & 2 & -3\\ 4 & -2 & 5 & 1\\ -5 & 0 & -4 & 0\\ 9 & 7 & 8 & -7 \end(array) \right)$.
We have a square matrix of the fourth order. Let us immediately note that the rank of this matrix does not exceed 4, i.e. $\rang A≤ 4$. Let's start finding the rank of the matrix.
Among the first-order minors (i.e., among the elements of the matrix $A$) there is at least one that is not equal to zero, therefore $\rang A≥ 1$. Let's move on to checking second-order minors. For example, at the intersection of rows No. 2, No. 3 and columns No. 1 and No. 2, we obtain the following second-order minor: $\left| \begin(array) (cc) 4 & -2 \\ -5 & 0 \end(array) \right|$. Let's calculate it:
$$\left| \begin(array) (cc) 4 & -2 \\ -5 & 0 \end(array) \right|=0-10=-10. $$
Among the second-order minors there is at least one that is not equal to zero, so $\rang A≥ 2$.
Let's move on to third-order minors. Let's find, for example, a minor whose elements are located at the intersection of rows No. 1, No. 3, No. 4 and columns No. 1, No. 2, No. 4:
$$\left | \begin(array) (cccc) -1 & 3 & -3\\ -5 & 0 & 0\\ 9 & 7 & -7 \end(array) \right|=105-105=0. $$
Since this third-order minor turned out to be equal to zero, it is necessary to investigate another third-order minor. Either all of them will be equal to zero (then the rank will be equal to 2), or among them there will be at least one that is not equal to zero (then we will begin to study fourth-order minors). Let's consider a third-order minor, the elements of which are located at the intersection of rows No. 2, No. 3, No. 4 and columns No. 2, No. 3, No. 4:
$$\left| \begin(array) (ccc) -2 & 5 & 1\\ 0 & -4 & 0\\ 7 & 8 & -7 \end(array) \right|=-28. $$
Among the third-order minors there is at least one non-zero, so $\rang A≥ 3$. Let's move on to checking fourth-order minors.
Any fourth-order minor is located at the intersection of four rows and four columns of the matrix $A$. In other words, the fourth-order minor is the determinant of the matrix $A$, since this matrix contains 4 rows and 4 columns. The determinant of this matrix was calculated in example No. 2 of the topic “Reducing the order of the determinant. Decomposing the determinant in a row (column)”, so let’s just take the finished result:
$$\left| \begin(array) (cccc) -1 & 3 & 2 & -3\\ 4 & -2 & 5 & 1\\ -5 & 0 & -4 & 0\\ 9 & 7 & 8 & -7 \end (array)\right|=86. $$
So the fourth order minor is not equal to zero. We can no longer form minors of the fifth order. Conclusion: the highest order of minors, among which there is at least one non-zero, is 4. Result: $\rang A=4$.
Answer: $\rang A=4$.
Example No. 3
Find the rank of the matrix $A=\left(\begin(array) (cccc) -1 & 0 & 2 & -3\\ 4 & -2 & 5 & 1\\ 7 & -4 & 0 & -5 \end( array) \right)$.
Let us immediately note that this matrix contains 3 rows and 4 columns, so $\rang A≤ 3$. In the previous examples, we began the process of finding the rank by considering minors of the smallest (first) order. Here we will try to immediately check the minors of the highest possible order. For the matrix $A$ these are the third order minors. Let's consider a third-order minor, the elements of which lie at the intersection of rows No. 1, No. 2, No. 3 and columns No. 2, No. 3, No. 4:
$$\left| \begin(array) (ccc) 0 & 2 & -3\\ -2 & 5 & 1\\ -4 & 0 & -5 \end(array) \right|=-8-60-20=-88. $$
So, the highest order of minors, among which there is at least one that is not equal to zero, is 3. Therefore, the rank of the matrix is 3, i.e. $\rang A=3$.
Answer: $\rang A=3$.
In general, finding the rank of a matrix by definition is, in the general case, a rather labor-intensive task. For example, a relatively small matrix of size $5\times 4$ has 60 second-order minors. And even if 59 of them are equal to zero, then the 60th minor may turn out to be non-zero. Then you will have to study third-order minors, of which this matrix has 40 pieces. Usually they try to use less cumbersome methods, such as the method of bordering minors or the method of equivalent transformations.
Rows (columns). Several rows (columns) are said to be linearly independent if none of them can be expressed linearly in terms of the others. The rank of the row system is always equal to the rank of the column system, and this number is called the rank of the matrix.
The rank of a matrix is the highest of the orders of all possible non-zero minors of this matrix. The rank of a zero matrix of any size is zero. If all second-order minors are zero, then the rank is one, etc.
Matrix rank - image dimension dim (im (A)) (\displaystyle \dim(\operatorname (im) (A))) linear operator to which the matrix corresponds.
Typically the rank of the matrix A (\displaystyle A) denoted by rang A (\displaystyle \operatorname (rang) A), r A (\displaystyle \operatorname (r) A), rg A (\displaystyle \operatorname (rg) A) or rank A (\displaystyle \operatorname (rank) A). The last option is typical for the English language, while the first two are for German, French and a number of other languages.
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Let be a rectangular matrix.
Then, by definition, the rank of the matrix A (\displaystyle A) is:
Theorem (about the correctness of determining ranks). Let all the minors of the matrix A m × n (\displaystyle A_(m\times n)) order k (\displaystyle k) are equal to zero ( M k = 0 (\displaystyle M_(k)=0)). Then ∀ M k + 1 = 0 (\displaystyle \forall M_(k+1)=0), if they exist.
Related definitions
Properties
- Theorem (about the basis minor): Let r = rang A , M r (\displaystyle r=\operatorname (rang) A,M_(r))- basis minor of the matrix A (\displaystyle A), Then:
- Consequences:
- Theorem (about rank invariance under elementary transformations): Let us introduce a notation for matrices obtained from each other by elementary transformations. Then the following statement is true: If A ∼ B (\displaystyle A\sim B), then their ranks are equal.
- Kronecker-Capelli theorem: A system of linear algebraic equations is consistent if and only if the rank of its main matrix is equal to the rank of its extended matrix. In particular:
- The number of main variables of the system is equal to the rank of the system.
- A consistent system will be defined (its solution is unique) if the rank of the system is equal to the number of all its variables.
- Sylvester's inequality: If A And B size matrices m x n And n x k, That
This is a special case of the following inequality.
- Frobenius' inequality: If AB, BC, ABC are correctly defined, then
Linear transformation and matrix rank
Let A (\displaystyle A)- size matrix m × n (\displaystyle m\times n) over the field C (\displaystyle C)(or R (\displaystyle R)). Let T (\displaystyle T)- linear transformation corresponding A (\displaystyle A) on a standard basis; it means that T (x) = A x (\displaystyle T(x)=Ax). Matrix rank A (\displaystyle A) is the dimension of the transformation range T (\displaystyle T).
Methods
There are several methods for finding the rank of a matrix:
- Elementary transformation method
- Bordering minor method
Any matrix A order m×n can be considered as a collection m string vectors or n column vectors.
Rank matrices A order m×n is the maximum number of linearly independent column vectors or row vectors.
If the matrix rank A equals r, then it is written:
Finding the rank of a matrix
Let A arbitrary order matrix m× n. To find the rank of a matrix A We apply the Gaussian elimination method to it.
Note that if at some stage of elimination the leading element is equal to zero, then we swap this line with the line in which the leading element is different from zero. If it turns out that there is no such line, then move on to the next column, etc.
After the forward Gaussian elimination process, we obtain a matrix whose elements under the main diagonal are equal to zero. In addition, there may be zero row vectors.
The number of non-zero row vectors will be the rank of the matrix A.
Let's look at all this with simple examples.
Example 1.
Multiplying the first line by 4 and adding to the second line and multiplying the first line by 2 and adding to the third line we have:
Multiply the second line by -1 and add it to the third line:
We received two non-zero rows and, therefore, the rank of the matrix is 2.
Example 2.
Let's find the rank of the following matrix:
Multiply the first line by -2 and add it to the second line. Similarly, we reset the elements of the third and fourth rows of the first column:
Let's reset the elements of the third and fourth rows of the second column by adding the corresponding rows to the second row multiplied by the number -1.
A number r is called the rank of matrix A if:
1) in the matrix A there is a minor of order r, different from zero;
2) all minors of order (r+1) and higher, if they exist, are equal to zero.
Otherwise, the rank of a matrix is the highest minor order other than zero.
Designations: rangA, r A or r.
From the definition it follows that r is a positive integer. For a null matrix, the rank is considered to be zero.Purpose of the service. The online calculator is designed to find matrix rank. In this case, the solution is saved in Word and Excel format. see example solution.
Instructions. Select the matrix dimension, click Next.
Definition . Let a matrix of rank r be given. Any minor of a matrix that is different from zero and has order r is called basic, and the rows and columns of its components are called basic rows and columns.
According to this definition, a matrix A can have several basis minors.The rank of the identity matrix E is n (the number of rows).
Example 1. Given two matrices,
and their minors 
, 
. Which of them can be taken as the basic one?
Solution. Minor M 1 =0, so it cannot be a basis for any of the matrices. Minor M 2 =-9≠0 and has order 2, which means it can be taken as the basis of matrices A or / and B, provided that they have ranks equal to 2. Since detB=0 (as a determinant with two proportional columns), then rangB=2 and M 2 can be taken as the basis minor of matrix B. The rank of matrix A is 3, due to the fact that detA=-27≠0 and, therefore, the order the basis minor of this matrix must be equal to 3, that is, M 2 is not a basis for the matrix A. Note that the matrix A has a single basis minor, equal to the determinant of the matrix A.Theorem (about the basis minor). Any row (column) of a matrix is a linear combination of its basis rows (columns).
Corollaries from the theorem.- Every (r+1) column (row) matrix of rank r is linearly dependent.
- If the rank of a matrix is less than the number of its rows (columns), then its rows (columns) are linearly dependent. If rangA is equal to the number of its rows (columns), then the rows (columns) are linearly independent.
- The determinant of a matrix A is equal to zero if and only if its rows (columns) are linearly dependent.
- If you add another row (column) to a row (column) of a matrix, multiplied by any number other than zero, then the rank of the matrix will not change.
- If you cross out a row (column) in a matrix, which is a linear combination of other rows (columns), then the rank of the matrix will not change.
- The rank of a matrix is equal to the maximum number of its linearly independent rows (columns).
- The maximum number of linearly independent rows is the same as the maximum number of linearly independent columns.
Example 2. Find the rank of a matrix
.
Solution. Based on the definition of the matrix rank, we will look for a minor of the highest order, different from zero. First, let's transform the matrix to a simpler form. To do this, multiply the first row of the matrix by (-2) and add it to the second, then multiply it by (-1) and add it to the third.Definition. Matrix rank is the maximum number of linearly independent rows considered as vectors.
Theorem 1 on the rank of the matrix. Matrix rank is called the maximum order of a nonzero minor of a matrix.
We already discussed the concept of a minor in the lesson on determinants, and now we will generalize it. Let's take a certain number of rows and a certain number of columns in the matrix, and this “how much” should be less than the number of rows and columns of the matrix, and for rows and columns this “how many” should be the same number. Then at the intersection of how many rows and how many columns there will be a matrix of lower order than our original matrix. The determinant is a matrix and will be a minor of the kth order if the mentioned “some” (the number of rows and columns) is denoted by k.
Definition. Minor ( r+1)th order, within which the chosen minor lies r-th order is called bordering for a given minor.
The two most commonly used methods are finding the rank of the matrix. This way of bordering minors And method of elementary transformations(Gauss method).
When using the bordering minors method, the following theorem is used.
Theorem 2 on the rank of the matrix. If a minor can be composed from matrix elements r th order, not equal to zero, then the rank of the matrix is equal to r.
When using the elementary transformation method, the following property is used:
If, through elementary transformations, a trapezoidal matrix is obtained that is equivalent to the original one, then rank of this matrix is the number of lines in it other than lines consisting entirely of zeros.
Finding the rank of a matrix using the method of bordering minors
An enclosing minor is a minor of a higher order relative to the given one if this minor of a higher order contains the given minor.
For example, given the matrix
Let's take a minor
The bordering minors will be:
Algorithm for finding the rank of a matrix next.
1. Find minors of the second order that are not equal to zero. If all second-order minors are equal to zero, then the rank of the matrix will be equal to one ( r =1 ).
2. If there is at least one minor of the second order that is not equal to zero, then we compose the bordering minors of the third order. If all bordering minors of the third order are equal to zero, then the rank of the matrix is equal to two ( r =2 ).
3. If at least one of the bordering minors of the third order is not equal to zero, then we compose the bordering minors. If all the bordering minors of the fourth order are equal to zero, then the rank of the matrix is equal to three ( r =2 ).
4. Continue this way as long as the matrix size allows.
Example 1. Find the rank of a matrix
.Solution. Minor of the second order
.Let's border it. There will be four bordering minors:
,
,Thus, all bordering minors of the third order are equal to zero, therefore, the rank of this matrix is equal to two ( r =2 ).
Example 2. Find the rank of a matrix
Solution. The rank of this matrix is equal to 1, since all the second-order minors of this matrix are equal to zero (in this, as in the cases of bordering minors in the two following examples, dear students are invited to verify for themselves, perhaps using the rules for calculating determinants), and among the first-order minors , that is, among the elements of the matrix, there are non-zero ones.
Example 3. Find the rank of a matrix
Solution. The second order minor of this matrix is , and all third order minors of this matrix are equal to zero. Therefore, the rank of this matrix is two.
Example 4. Find the rank of a matrix
Solution. The rank of this matrix is 3, since the only third-order minor of this matrix is 3.
Finding the rank of a matrix using the method of elementary transformations (Gauss method)
Already in example 1 it is clear that the task of determining the rank of a matrix using the method of bordering minors requires the calculation of a large number of determinants. There is, however, a way to reduce the amount of computation to a minimum. This method is based on the use of elementary matrix transformations and is also called the Gauss method.
The following operations are understood as elementary matrix transformations:
1) multiplying any row or column of a matrix by a number other than zero;
2) adding to the elements of any row or column of the matrix the corresponding elements of another row or column, multiplied by the same number;
3) swapping two rows or columns of the matrix;
4) removing “null” rows, that is, those whose elements are all equal to zero;
5) deleting all proportional lines except one.
Theorem. During an elementary transformation, the rank of the matrix does not change. In other words, if we use elementary transformations from the matrix A went to the matrix B, That .
