“If you want to learn how to swim, then boldly enter the water, and if you want to learn to solve problems, then solve them.»
D. Poya (1887-1985)
(Mathematician. Made a great contribution to the popularization of mathematics. Wrote several books on how to solve problems and how to teach how to solve problems.)
Consider the matrix
Let's highlight in it k-rows And k-columns (k≤(min(m,n))). From the elements at the intersection of the selected rows and columns, we will compose the determinant k-th order. All such determinants are called minors of this matrix.
Consider all possible minors of the matrix BUT, which are different from zero.
Matrix rank BUT is the largest order of the non-zero minor of this matrix.
If all elements of a matrix are equal to zero, then the rank of this matrix is taken equal to zero.
A minor whose order determines the rank of a matrix is called basic.
A matrix can have multiple basis minors.
Matrix rank BUT denoted r(A). If r(A)=r(B), then the matrices BUT And IN called equivalent. Write A∼B.
Matrix rank properties:
- Transposing a matrix does not change its rank.
- If you cross out the zero row (column) from the matrix, then the rank of the matrix will not change.
- The rank of a matrix does not change under elementary matrix transformations.
Elementary transformations mean:
- Permutation of matrix rows;
- Multiplying any string by a non-zero number;
- Adding to the elements of one row the corresponding elements of another row, multiplied by an arbitrary number.
When calculating the rank of a matrix, elementary transformations, the method of reducing a matrix to a stepped form, the method of bordering minors can be used.
Method for reducing a matrix to a stepwise The point is that with the help of elementary transformations, this matrix is reduced to a step matrix.
The matrix is called stepped , if the first non-zero element in each of its rows is to the right than in the previous one (i.e. steps are obtained, the height of each step must be equal to one).
Examples of step matrices:
Examples of non-step matrices:
EXAMPLE: Find the rank of a matrix:
SOLUTION:
Let us reduce this matrix to a step matrix with the help of elementary transformations.
1. Swap the first and third lines.
2. We get zeros under one in the first column.
Adding to the second line the first multiplied by (-3), to the third - the first multiplied by (-5), to the fourth - the first multiplied by (-3), we get
In order to make it clearer where else you need to get zeros, let's draw steps in the matrix. (The matrix will be stepped if there are zeros everywhere under the steps)
3. Adding to the third row the second, multiplied by (-1), to the fourth - the second, multiplied by (-1), we get zeros under the steps in the second column.
If we draw steps again, we will see that the matrix is stepped.
Her rank is r=3(the number of rows of the step matrix, in each of which at least one element is different from zero). Therefore, the rank of this matrix r=3.
The solution can be written like this:
(Roman numerals indicate line numbers)
Answer: r=3.
Minor order k+1, which contains the minor of the order k called edging minor.
Fringing Minor Method is based on the fact that the rank of a given matrix is equal to the order of such a minor of this matrix, which is different from zero, and all the minors bordering it are equal to zero.
Definition. Matrix rank is the maximum number of linearly independent rows considered as vectors.
Theorem 1 on the rank of a matrix. Matrix rank is the maximum order of a non-zero minor of a matrix.
We have already discussed the concept of a minor in the lesson on determinants, and now we will generalize it. Let's take some rows and some columns in the matrix, and this "something" should be less than the number of rows and columns of the matrix, and for rows and columns this "something" should be the same number. Then at the intersection of how many rows and how many columns there will be a matrix of a smaller order than our original matrix. The determinant of this matrix will be a k-th order minor if the mentioned "something" (the number of rows and columns) is denoted by k.
Definition. Minor ( r+1)-th order, inside which lies the chosen minor r-th order, is called bordering for the given minor.
The two most commonly used methods finding the rank of a matrix. This way of fringing minors And method of elementary transformations(by the Gauss method).
The method of bordering minors uses the following theorem.
Theorem 2 on the rank of a matrix. If it is possible to compose a minor from the elements of the matrix r th order, which is not equal to zero, then the rank of the matrix is equal to r.
With the method of elementary transformations, the following property is used:
If a trapezoidal matrix equivalent to the original one is obtained by elementary transformations, then the rank of this matrix is the number of lines in it except for lines consisting entirely of zeros.
Finding the rank of a matrix by the method of bordering minors
A bordering minor is a minor of a higher order in relation 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
edging will be such minors:
Algorithm for finding the rank of a matrix next.
1. We 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 exists at least one second-order minor that is not equal to zero, then we compose bordering third-order minors. If all third-order bordering minors are zero, then the rank of the matrix is 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 minors bordering it. If all bordering fourth-order minors are zero, then the rank of the matrix is three ( r =2 ).
4. Continue as long as the size of the matrix allows.
Example 1 Find the rank of a matrix
.
Solution. Minor of the second order 
.
We frame it. There will be four bordering minors:
,
,
Thus, all bordering third-order minors are equal to zero, therefore, the rank of this matrix is two ( r =2 ).
Example 2 Find the rank of a matrix
Solution. The rank of this matrix is 1, since all second-order minors of this matrix are equal to zero (in this, as in the cases of bordering minors in the next two examples, dear students are invited to verify for themselves, perhaps using the rules for calculating determinants), and among first-order minors , that is, among the elements of the matrix, there are not equal to zero.
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 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 because the only third order minor of this matrix is 3.
Finding the rank of a matrix by the method of elementary transformations (by the Gauss method)
Already in Example 1, it can be seen that the problem of determining the rank of a matrix by 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.
Elementary transformations of a matrix mean the following operations:
1) multiplication of any row or any column of the matrix by a number other than zero;
2) adding to the elements of any row or any column of the matrix the corresponding elements of another row or column, multiplied by the same number;
3) swapping two rows or columns of a matrix;
4) removal of "null" rows, that is, those, all elements of which are equal to zero;
5) deletion of all proportional lines, except for one.
Theorem. The elementary transformation does not change the rank of the matrix. In other words, if we use elementary transformations from the matrix A go to matrix B, then .
Determining the rank of a matrix
Consider a matrix \(A\) of type \((m,n)\). Let, for definiteness, \(m \leq n\). Take \(m\) rows and choose \(m\) columns of the matrix \(A\), at the intersection of these rows and columns we get a square matrix of order \(m\), whose determinant is called minor order \(m\) matrices \(A\). If this minor is different from 0, it is called basic minor and say that the rank of the matrix \(A\) is \(m\). If this determinant is equal to 0, then other \(m\) columns are chosen, at their intersection there are elements that form another minor of order \(m\). If the minor is 0, we continue the procedure. If among all possible minors of order \(m\) there are no non-zero ones, we select \(m-1\) rows and columns from the matrix \(A\), at their intersection a square matrix of order \(m-1\) appears , its determinant is called the order minor \(m-1\) of the original matrix. Continuing the procedure, we look for a non-zero minor, going through all possible minors, lowering their order.
Definition.
A non-zero minor of a given matrix of the highest order is called basic minor of the original matrix, its order is called rank matrices \(A\), rows and columns, at the intersection of which there is a basic minor, are called basic rows and columns. The rank of a matrix is denoted by \(rang(A)\).
Simple properties of the rank of a matrix follow from this definition: it is an integer, and the rank of a non-zero matrix satisfies the inequalities: \(1 \leq rang(A) \leq \min(m,n)\).
How will the rank of the matrix change if a row is crossed out? Add some line?
Check Answer
1) The rank can decrease by 1.
2) The rank can increase by 1.
Linear dependence and linear independence of matrix columns
Let \(A\) be a matrix of type \((m,n)\). Consider the columns of the matrix \(A\) - these are columns of \(m\) numbers each. Let's denote them \(A_1,A_2,...,A_n\). Let \(c_1,c_2,...,c_n\) be some numbers.
Definition.
The column \[ D=c_1A_1+c_2A_2+...+c_nA_n = \sum _(m=1)^nc_mA_m \] is called the linear combination of the columns \(A_1,A_2,...,A_n\), numbers \(c_1,c_2 ,...,c_n\) are called the coefficients of this linear combination.
Definition.
Let \(p\) columns \(A_1, A_2, ..., A_p\) be given. If there are numbers \(c_1,c_2,...,c_p\) such that
1. not all of these numbers are zero,
2. the linear combination \(c_1A_1+c_2A_2+...+c_pA_p =\sum _(m=1)^pc_mA_m\) is equal to the zero column (that is, the column, all elements of which are zeros), then we say that the columns \( A_1, A_2, ..., A_p\) are linearly dependent. If there are no such numbers \(c_1,c_2,...,c_n\) for a given set of columns, the columns are said to be linearly independent.
Example. Consider 2-columns
\[ A_1=\left(\begin(array)(c) 1 \\ 0 \end(array) \right), A_2=\left(\begin(array)(c) 0 \\ 1 \end(array) \right), \] then for any numbers \(c_1,c_2\) we have: \[ c_1A_1+c_2A_2=c_1\left(\begin(array)(c) 1 \\ 0 \end(array) \right)+ c_2\left(\begin(array)(c) 0 \\ 1 \end(array) \right)=\left(\begin(array)(c) c_1 \\ c_2 \end(array) \right). \]
This linear combination is equal to the zero column if and only if both numbers \(c_1,c_2\) are equal to zero. Thus, these columns are linearly independent.
Statement. For columns to be linearly dependent, it is necessary and sufficient that one of them be a linear combination of the others.
Let the columns \(A_1,A_2,...,A_m\) be linearly dependent, i.e. for some constants \(\lambda _1, \lambda _2,...,\lambda _m\), not all of which are 0, the following is executed: \[ \sum _(k=1)^m\lambda _kA_k=0 \ ] (on the right side - zero column). Let, for example, \(\lambda _1 \neq 0\). Then \[ A_1=\sum _(k=2)^mc_kA_k, \quad c_k=-\lambda _k/\lambda _1, \quad \quad (15) \] i.e. the first column is a linear combination of the rest.
Basis minor theorem
Theorem.For any non-zero matrix \(A\) the following is true:
1. Basic columns are linearly independent.
2. Any column of a matrix is a linear combination of its basic columns.
(The same is true for strings).
Let, for definiteness, \((m,n)\) be the type of the matrix \(A\), \(rang(A)=r \leq n\) and the basis minor is located in the first \(r\) rows and columns matrices \(A\). Let \(s\) be any number between 1 and \(m\), \(k\) be any number between 1 and \(n\). Consider a minor of the following form: \[ D=\left| \begin(array)(ccccc) a_(11) & a_(12) & \ldots & a_(1r) & a_(1s) \\ a_(21) & a_(22) & \ldots & a_(2r) & a_(2s) \\ \dots &\ldots & \ldots & \ldots & \ldots \\ a_(r1) & a_(r2) & \ldots & a_(rr) & a_(rs) \\ a_(k1) & a_(k2) & \ldots & a_(kr) & a_(ks) \\ \end(array) \right| , \] i.e. we have assigned the \(s-\)th column and the \(k-\)th row to the basic minor. By definition of the matrix rank, this determinant is equal to zero (if we chose \(s\leq r\) or \(k \leq r\) , then this minor has 2 identical columns or 2 identical rows, if \(s>r\) and \(k>r\) - by definition of rank, the minor of size greater than \(r\) vanishes). Expanding this determinant over the last line, we get: \[ a_(k1)A_(k1)+a_(k2)A_(k2)+....+a_(kr)A_(kr)+a_(ks)A_(ks )=0. \quad \quad(16) \]
Here the numbers \(A_(kp)\) are the algebraic complements of the elements from the bottom row \(D\). Their values do not depend on \(k\), because are formed using elements from the first \(r\) lines. In this case, \(A_(ks)\) is a basic minor other than 0. Denote \(A_(k1)=c_1,A_(k2)=c_2,...,A_(ks)=c_s \neq 0 \). Let us rewrite (16) in new notation: \[ c_1a_(k1)+c_2a_(k2)+...+c_ra_(kr)+c_sa_(ks)=0, \] or, dividing by \(c_s\), \[ a_(ks)=\lambda_1a_(k1)+\lambda_2a_(k2)+...+\lambda_ra_(kr), \quad \lambda _p=-c_p/c_s. \] This equality holds for any value \(k\), so \[ a_(1s)=\lambda_1a_(11)+\lambda_2a_(12)+...+\lambda_ra_(1r), \] \[ a_ (2s)=\lambda_1a_(21)+\lambda_2a_(22)+...+\lambda_ra_(2r), \] \[ .................... .................................... \] \[ a_(ms)=\lambda_1a_(m1) +\lambda_2a_(m2)+...+\lambda_ra_(mr). \] So the \(s-\)th column is a linear combination of the first \(r\) columns. The theorem has been proven.
Comment.
The basis minor theorem implies that the rank of a matrix is equal to the number of its linearly independent columns (which is equal to the number of linearly independent rows).
Consequence 1.
If the determinant is zero, then it has a column that is a linear combination of the rest of the columns.
Consequence 2.
If the rank of a matrix is less than the number of columns, then the columns of the matrix are linearly dependent.
Calculating the rank of a matrix and finding the basis minor
Some transformations of a matrix do not change its rank. Such transformations can be called elementary. The corresponding facts can be easily verified using the properties of the determinants and the definition of the rank of a matrix.
1. Rearranging columns.
2. Multiplication of elements of any column by a non-zero factor.
3. Adding to a column of any other column, multiplied by an arbitrary number.
4. Crossing out the zero column.
The same is true for strings.
With the help of these transformations, the matrix can be transformed to the so-called "trapezoidal" form - a matrix, under the main diagonal of which there are only zeros. For a "trapezoid" matrix, the rank is the number of non-zero elements on the main diagonal, and the basis minor is the minor whose diagonal matches the set of non-zero elements on the main diagonal of the transformed matrix.
Example. Consider the matrix
\[ A=\left(\begin(array)(cccc) 2 &1 & 11 & 2 \\ 1 & 0 & 4 & -1 \\ 11 & 4 & 56 & 5 \\ 2 & -1 & 5 & - 6 \end(array)\right). \] We will transform it using the above transformations. \[ A=\left(\begin(array)(cccc) 2 &1 & 11 & 2 \\ 1 & 0 & 4 & -1 \\ 11 & 4 & 56 & 5 \\ 2 & -1 & 5 & - 6 \end(array) \right) \mapsto \left(\begin(array)(cccc) 1 & 0 & 4 & -1 \\ 2 & 1 & 11 & 2 \\ 11 & 4 & 56 & 5 \\ 2 & -1 & 5 & -6 \end(array) \right) \mapsto \left(\begin(array)(cccc) 1 & 0 & 4 & -1 \\ 0 & 1 & 3 & 4 \\ 0 & 4 & 12 & 16 \\ 0 & -1 & -3 & -4 \end(array) \right) \mapsto \] \[ \left(\begin(array)(cccc) 1 & 0 & 4 & - 1 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end(array) \right)\mapsto \left(\begin(array)(cccc) 1 & 0 & 4 & -1 \\ 0 & 1 & 3 & 4 \end(array)\right). \]
Here we consistently take the following steps: 1) rearrange the second row up, 2) subtract the first row from the rest with a suitable factor, 3) subtract the second row from the third 4 times, add the second row to the fourth, 4) cross out the zero rows - the third and fourth . Our final matrix has acquired the desired shape: there are non-zero numbers on the main diagonal, zeros under the main diagonal. After that, the procedure stops and the number of nonzero elements on the main diagonal is equal to the rank of the matrix. In this case, the basic minor is the first two rows and the first two columns. At their intersection there is a matrix of order 2 with a nonzero determinant. At the same time, returning along the chain of transformations in the opposite direction, one can trace where this or that row (this or that column) came from in the final matrix, i.e. determine the basic rows and columns in the original matrix. In this case, the first two rows and the first two columns form the basis minor.
To work with the concept of the rank of a matrix, we 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 a matrix precisely through minors.
Matrix rank name the maximum order of its minors, among which there is at least one that is not equal to zero.
Equivalent matrices are matrices whose ranks are equal to each other.
Let's explain in more detail. Suppose there is at least one among second-order minors that is different from zero. And all minors, the order of which 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, the order of which 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 set equal to zero, $\rang O=0$. Let me remind you that in order to form a matrix minor, it is required 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. it contains 5 rows and 4 columns), then the maximum order of its minors is four. It will no longer be possible to form fifth-order minors, since they will require 5 columns (and we have only 4). This means that the rank of the matrix $F$ cannot be greater than four, i.e. $\rang F≤4$.
In a more general form, the above means that if the matrix contains $m$ rows and $n$ columns, then its rank cannot exceed the smallest of the numbers $m$ and $n$, i.e. $\rang A≤\min(m,n)$.
In principle, the method of finding it follows from the very definition of the rank. 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. with first-order minors of some matrix $A$.
- If all first-order minors (that is, 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 $\rang A≥ 1$. We pass to the verification of minors of the second order.
- 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 $\rang A≥ 2$. We pass to the verification of minors of the third order.
- If all third-order minors are equal to zero, then $\rang A=2$. If among the minors of the third order there is at least one that is not equal to zero, then $\rang A≥ 3$. Let's move on to checking the minors of the fourth order.
- If all fourth-order minors are equal to zero, then $\rang A=3$. If there is at least one non-zero minor of the fourth order, then $\rang A≥ 4$. We pass to the verification of minors of the fifth order, and so on.
What awaits us at the end of this procedure? It is possible that among the minors of the kth order there is at least one that is different from zero, and all the minors of the (k + 1)th order 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 minors of the kth order there will be at least one that is not equal to zero, and the minors of the (k + 1)th order cannot be formed. In this case, the rank of the matrix is also equal to k. Shortly speaking, the order of the last composed non-zero minor and 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 illustrated clearly. Once again, I emphasize that in the examples of this topic, we will find the rank of matrices using only the definition of the rank. Other methods (calculation of the rank of a matrix by the method of bordering minors, calculation of the rank of a matrix by the method of elementary transformations) are considered in the following topics.
By the way, it is not at all necessary to start the procedure for finding the rank from minors of the smallest order, as was done in examples No. 1 and No. 2. You can immediately go to minors of higher orders (see example No. 3).
Example #1
Find the rank of a 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, 3 is the minimum, so the rank of the matrix $A$ is at most 3, i.e. $\rank A≤ 3$. And this inequality is obvious, since we can no longer form minors of the fourth order - they need 4 rows, and we have only 3. Let's proceed directly to the process of finding the rank of a given matrix.
Among the minors of the first order (that is, 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 there is at least one non-zero among the first-order minors, we conclude that $\rang A≥ 1$ and proceed to check the second-order minors.
Let's start exploring minors of the second order. For example, at the intersection of rows #1, #2 and columns #1, #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 #3 in the property 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 minor of the second order we checked turned out to be equal to zero. What does it say? About the need to further check second-order minors. Either they all turn out to be zero (and then the rank will be equal to 1), or among them there is at least one minor that is different from zero. Let's try to make a better choice by writing a second-order minor whose elements are located at the intersection of rows #1, #2 and columns #1 and #5: $\left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array)\right|$. Let's find the value of this minor of the second order:
$$ \left|\begin(array)(cc) 5 & 2 \\ 7 & 3 \end(array) \right|=5\cdot 3-2\cdot 7=1. $$
This minor is not equal to zero. Conclusion: among the minors of the second order there is at least one other than zero. Hence $\rank A≥ 2$. It is necessary to proceed to the study of minors of the third order.
If for the formation of minors of the third order we choose column #2 or column #4, then such minors will be equal to zero (because they will contain a zero column). It remains to check only one minor of the third order, 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 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 the second order. Conclusion: the maximum order of minors, among which there is at least one other than zero, is equal to 2. Therefore, $\rang A=2$.
Answer: $\rank A=2$.
Example #2
Find the rank of a 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. We note right away that the rank of this matrix does not exceed 4, i.e. $\rank A≤ 4$. Let's start finding the rank of a matrix.
Among the minors of the first order (that is, among the elements of the matrix $A$) there is at least one that is not equal to zero, so $\rang A≥ 1$. We pass to the verification of minors of the second order. For example, at the intersection of rows No. 2, No. 3 and columns No. 1 and No. 2, we get the following minor of the second order: $\left| \begin(array) (cc) 4 & -2 \\ -5 & 0 \end(array) \right|$. 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 minors of the third order. 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 minors of the fourth order). Consider a third-order minor whose elements are located at the intersection of rows No. 2, No. 3, No. 4 and columns No. 2, No. 3, No. 4:
$$ \left| \begin(array) (ccc) -2 & 5 & 1\\ 0 & -4 & 0\\ 7 & 8 & -7 \end(array) \right|=-28. $$
There is at least one non-zero minor among third-order minors, so $\rang A≥ 3$. Let's move on to checking the minors of the fourth order.
Any minor of the fourth order 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 just 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. Decomposition of 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 other than zero, is 4. Result: $\rang A=4$.
Answer: $\rank A=4$.
Example #3
Find the rank of a matrix $A=\left(\begin(array) (cccc) -1 & 0 & 2 & -3\\ 4 & -2 & 5 & 1\\ 7 & -4 & 0 & -5 \end( array)\right)$.
Note right away that this matrix contains 3 rows and 4 columns, so $\rang A≤ 3$. In the previous examples, we started 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 third-order minors. Consider a third-order minor whose elements 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. $\rank A=3$.
Answer: $\rank A=3$.
In general, finding the rank of a matrix by definition is, in the general case, a rather time-consuming task. For example, a relatively small $5\times 4$ matrix 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 have to explore the third-order minors, of which this matrix has 40 pieces. Usually one tries to use less cumbersome methods, such as the method of bordering minors or the method of equivalent transformations.
The rank of a matrix is an important numerical characteristic. The most typical problem that requires finding the rank of a matrix is checking the compatibility of a system of linear algebraic equations. In this article, we will give the concept of the rank of a matrix and consider methods for finding it. For a better assimilation of the material, we will analyze in detail the solutions of several examples.
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Determination of the rank of a matrix and necessary additional concepts.
Before voicing the definition of the rank of a matrix, one should have a good understanding of the concept of a minor, and finding the minors of a matrix implies the ability to calculate the determinant. So we recommend, if necessary, to recall the theory of the article, methods for finding the matrix determinant, properties of the determinant.
Take a matrix A of order . Let k be some natural number not exceeding the smallest of the numbers m and n , that is, 
.
Definition.
Minor k-th order matrix A is the determinant of the square matrix of order , composed of the elements of the matrix A, which are in pre-selected k rows and k columns, and the location of the elements of the matrix A is preserved.
In other words, if we delete (p–k) rows and (n–k) columns in matrix A, and form a matrix from the remaining elements, keeping the arrangement of matrix elements A, then the determinant of the resulting matrix is a minor of order k of matrix A.
Let's look at the definition of a matrix minor using an example.
Consider the matrix 
.
Let us write down several first-order minors of this matrix. For example, if we choose the third row and the second column of the matrix A, then our choice corresponds to a first-order minor 
. In other words, to obtain this minor, we crossed out the first and second rows, as well as the first, third and fourth columns from the matrix A, and made up the determinant from the remaining element. If we choose the first row and the third column of the matrix A, then we get a minor 
.
Let us illustrate the procedure for obtaining the considered first-order minors 
And 
.
Thus, the first-order minors of a matrix are the matrix elements themselves.
Let us show several minors of the second order. Select two rows and two columns. For example, take the first and second rows and the third and fourth columns. With this choice, we have a second-order minor 
. This minor could also be formed by deleting the third row, first and second columns from matrix A.
Another second-order minor of matrix A is .
Let us illustrate the construction of these second-order minors 
And 
.
The third-order minors of the matrix A can be found similarly. Since there are only three rows in matrix A, we select them all. If we select the first three columns for these rows, then we get a minor of the third order 
It can also be constructed by deleting the last column of the matrix A.
Another third-order minor is 
obtained by deleting the third column of the matrix A.
Here is a drawing showing the construction of these third order minors 
And 
.
For a given matrix A, there are no minors of order higher than the third, since .
How many k-th order minors of the matrix A of order exist?
The number of order k minors can be calculated as , where 
And 
- the number of combinations from p to k and from n to k, respectively.
How to construct all minors of order k of matrix A of order p on n?
We need a set of matrix row numbers and a set of column numbers. Recording everything combinations of p elements by k(they will correspond to the selected rows of the matrix A when constructing a minor of order k). To each combination of row numbers, we sequentially add all combinations of n elements by k column numbers. These sets of combinations of row numbers and column numbers of matrix A will help to compose all minors of order k.
Let's take an example.
Example.
Find all the second order minors of the matrix.
Solution.
Since the order of the original matrix is 3 by 3, then the total second-order minors will be 
.
Let's write down all combinations of 3 to 2 row numbers of the matrix A: 1, 2; 1, 3 and 2, 3. All combinations of 3 by 2 column numbers are 1, 2 ; 1, 3 and 2, 3.
Take the first and second rows of matrix A. Selecting the first and second columns for these rows, the first and third columns, the second and third columns, we obtain, respectively, the minors 
For the first and third rows, with a similar choice of columns, we have 
It remains to add the first and second, first and third, second and third columns to the second and third rows: 
So, all nine minors of the second order of the matrix A are found.
Now we can move on to determining the rank of the matrix.
Definition.
Matrix rank is the highest order of the non-zero matrix minor.
The rank of matrix A is denoted as Rank(A) . You can also see the designations Rg(A) or Rang(A) .
From the definitions of the rank of a matrix and the minor of a matrix, we can conclude that the rank of a zero matrix is equal to zero, and the rank of a nonzero matrix is at least one.
Finding the rank of a matrix by definition.
So, the first method for finding the rank of a matrix is minor enumeration method. This method is based on determining the rank of the matrix.
Let us need to find the rank of a matrix A of order .
Briefly describe algorithm solution of this problem by the method of enumeration of minors.
If there is at least one matrix element other than zero, then the rank of the matrix is at least equal to one (since there is a first-order minor that is not equal to zero).
Next, we iterate over the minors of the second order. If all second-order minors are equal to zero, then the rank of the matrix is equal to one. If there exists at least one non-zero second-order minor, then we pass to the enumeration of third-order minors, and the rank of the matrix is at least equal to two.
Similarly, if all third-order minors are zero, then the rank of the matrix is two. If there is at least one non-zero third-order minor, then the rank of the matrix is at least three, and we proceed to the enumeration of fourth-order minors.
Note that the rank of a matrix cannot exceed the smallest of p and n.
Example.
Find the rank of a matrix 
.
Solution.
Since the matrix is non-zero, its rank is not less than one.
Minor of the second order 
is different from zero, therefore, the rank of the matrix A is at least two. We pass to the enumeration of minors of the third order. All of them 
things. 
All third-order minors are equal to zero. Therefore, the rank of the matrix is two.
Answer:
Rank(A) = 2 .
Finding the rank of a matrix by the method of fringing minors.
There are other methods for finding the rank of a matrix that allow you to get the result with less computational work.
One of these methods is fringing minor method.
Let's deal with the notion of a bordering minor.
It is said that the minor M ok of the (k+1)th order of the matrix A borders the minor M of order k of the matrix A if the matrix corresponding to the minor M ok “contains” the matrix corresponding to the minor M .
In other words, the matrix corresponding to the bordered minor M is obtained from the matrix corresponding to the bordering minor M OK by deleting the elements of one row and one column.
For example, consider the matrix 
and take a minor of the second order. Let's write down all bordering minors: 
The method of bordering minors is justified by the following theorem (we present its formulation without proof).
Theorem.
If all minors bordering the k-th order minor of a matrix A of order p by n are equal to zero, then all minors of order (k + 1) of matrix A are equal to zero.
Thus, to find the rank of a matrix, it is not necessary to enumerate all the minors that are bordering enough. The number of minors bordering the k-th order minor of the matrix A of order is found by the formula 
. Note that there are no more minors bordering the k-th order minor of the matrix A than there are (k + 1)-th order minors of the matrix A . Therefore, in most cases, using the method of bordering minors is more profitable than simply enumeration of all minors.
Let us proceed to finding the rank of a matrix by the method of fringing minors. Briefly describe algorithm this method.
If the matrix A is non-zero, then we take any element of the matrix A that is different from zero as a first-order minor. We consider its bordering minors. If they are all equal to zero, then the rank of the matrix is equal to one. If there is at least one non-zero bordering minor (its order is equal to two), then we pass to the consideration of its bordering minors. If they are all zero, then Rank(A) = 2 . If at least one bordering minor is nonzero (its order is equal to three), then we consider its bordering minors. Etc. As a result, Rank(A) = k if all bordering minors of the (k + 1)th order of the matrix A are equal to zero, or Rank(A) = min(p, n) if there exists a non-zero minor bordering the minor of order (min( p, n) – 1) .
Let's analyze the method of bordering minors for finding the rank of a matrix using an example.
Example.
Find the rank of a matrix 
by the bordering minors method.
Solution.
Since the element a 1 1 of the matrix A is non-zero, we take it as a first-order minor. Let's start searching for a bordering minor other than zero: 
A non-zero bordering second-order minor is found. Let us enumerate its bordering minors (their 
things): 
All minors bordering the second-order minor are equal to zero, therefore, the rank of matrix A is equal to two.
Answer:
Rank(A) = 2 .
Example.
Find the rank of a matrix 
with the help of bordering minors.
Solution.
As a non-zero minor of the first order, we take the element a 1 1 = 1 of the matrix A . Fringing it minor of the second order 
is not equal to zero. This minor is bordered by a minor of the third order 
. Since it is not equal to zero and there is no bordering minor for it, the rank of the matrix A is equal to three.
Answer:
Rank(A) = 3 .
Finding the rank using elementary transformations of the matrix (by the Gauss method).
Consider another way to find the rank of a matrix.
The following matrix transformations are called elementary:
- permutation of the rows (or columns) of the matrix;
- multiplication of all elements of any row (column) of the matrix by an arbitrary number k that is different from zero;
- adding to the elements of any row (column) the corresponding elements of another row (column) of the matrix, multiplied by an arbitrary number k.
Matrix B is called equivalent to matrix A, if B is obtained from A with the help of a finite number of elementary transformations. The equivalence of matrices is denoted by the symbol "~", that is, it is written A ~ B.
Finding the rank of a matrix using elementary matrix transformations is based on the statement: if matrix B is obtained from matrix A using a finite number of elementary transformations, then Rank(A) = Rank(B) .
The validity of this statement follows from the properties of the matrix determinant:
- When the rows (or columns) of a matrix are permuted, its determinant changes sign. If it is equal to zero, then when permuting the rows (columns), it remains equal to zero.
- When multiplying all elements of any row (column) of the matrix by an arbitrary number k different from zero, the determinant of the resulting matrix is equal to the determinant of the original matrix, multiplied by k. If the determinant of the original matrix is equal to zero, then after multiplying all the elements of any row or column by the number k, the determinant of the resulting matrix will also be equal to zero.
- Adding to the elements of a certain row (column) of the matrix the corresponding elements of another row (column) of the matrix, multiplied by some number k, does not change its determinant.
The essence of the method of elementary transformations is to bring the matrix, the rank of which we need to find, to a trapezoid (in a particular case, to an upper triangular one) using elementary transformations.
What is it for? The rank of matrices of this kind is very easy to find. It is equal to the number of rows containing at least one non-null element. And since the rank of the matrix does not change during elementary transformations, the resulting value will be the rank of the original matrix.
We give illustrations of matrices, one of which should be obtained after transformations. Their form depends on the order of the matrix.
These illustrations are templates to which we will transform the matrix A.
Let's describe method algorithm.
Suppose we need to find the rank of a non-zero matrix A of order (p can be equal to n).
So, . Let's multiply all the elements of the first row of matrix A by . In this case, we obtain an equivalent matrix, denote it A (1) : 
To the elements of the second row of the resulting matrix A (1), we add the corresponding elements of the first row, multiplied by . To the elements of the third row, add the corresponding elements of the first row, multiplied by . And so on up to the p-th line. We get an equivalent matrix, denote it A (2) : 
If all elements of the resulting matrix that are in rows from the second to the p-th are equal to zero, then the rank of this matrix is equal to one, and, consequently, the rank of the original matrix is equal to one.
If there is at least one non-zero element in the rows from the second to the p-th, then we continue to carry out transformations. Moreover, we act in exactly the same way, but only with the part of the matrix A marked in the figure (2) 
If , then we rearrange the rows and (or) columns of the matrix A (2) so that the "new" element becomes non-zero.
