Determine the rank of a matrix example. Find the rank of a matrix: methods and examples

This article will discuss such a concept as the rank of a matrix and the necessary additional concepts. We will give examples and proofs of finding the rank of a matrix, and also tell you what a matrix minor is and why it is so important.

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Matrix minor

To understand what the rank of a matrix is, it is necessary to understand such a concept as a matrix minor.

Definition 1

Minorkth order matrix - the determinant of a square matrix of the order k × k, which is composed of the elements of the matrix A, located in pre-selected k-rows and k-columns, while maintaining the position of the elements of the matrix A.

Simply put, if we delete (p-k) rows and (n-k) columns in matrix A, and make a matrix of those elements that remain, keeping the arrangement of the elements of matrix A, then the determinant of the resulting matrix is the order-k minor of matrix A.

It follows from the example that the first-order minors of the matrix A are the matrix elements themselves.

We can give several examples of minors of the 2nd order. Let's choose two rows and two columns. For example, 1st and 2nd row, 3rd and 4th column.

With this choice of elements, the minor of the second order will be - 1 3 0 2 = (- 1) × 2 - 3 × 0 = - 2

Another 2nd order minor of matrix A is 0 0 1 1 = 0

Let us provide illustrations of the construction of second-order minors of the matrix A:

The 3rd order minor is obtained by deleting the third column of matrix A:

0 0 3 1 1 2 - 1 - 4 0 = 0 × 1 × 0 + 0 × 2 × (- 1) + 3 × 1 × (- 4) - 3 × 1 × (- 1) - 0 × 1 × 0 - 0 × 2 × (- 4) = - 9

An illustration of how the 3rd order minor of matrix A is obtained:

For a given matrix, there are no minors higher than the 3rd order, because

k ≤ m i n (p , n) = m i n (3 , 4) = 3

How many k-th order minors are there for a matrix A of order p×n?

The number of minors is calculated using the following formula:

C p k × C n k , g e C p k = p ! k! (p - k) ! and C nk = n ! k! (n - k) ! - the number of combinations from p to k, from n to k, respectively.

After we have decided what the minors of the matrix A are, we can proceed to determining the rank of the matrix A.

Matrix rank: methods of finding

Definition 2

Matrix rank - the highest order of the matrix, other than zero.

Designation 1

Rank (A), Rg(A), Rang(A).

From the definition of the rank of a matrix and the minor of a matrix, it becomes clear that the rank of a zero matrix is equal to zero, and the rank of a non-zero matrix is different from zero.

Finding the rank of a matrix by definition

Definition 3

Minor enumeration method - a method based on determining the rank of a matrix.

Algorithm of actions by enumeration of minors :

It is necessary to find the rank of the matrix A of order p× n. If there is at least one non-zero element, then the rank of the matrix is at least equal to one ( because is a 1st order minor that is not equal to zero).

Then follows the enumeration of minors of the 2nd order. If all 2nd order minors are equal to zero, then the rank is equal to one. If there is at least one non-zero minor of the 2nd order, it is necessary to go to the enumeration of minors of the 3rd order, and the rank of the matrix, in this case, will be at least two.

Let's do the same with the rank of the 3rd order: if all the minors of the matrix are equal to zero, then the rank will be equal to two. If there is at least one non-zero third-order minor, then the rank of the matrix is at least three. And so on, by analogy.

Example 2

Find the rank of a matrix:

A \u003d - 1 1 - 1 - 2 0 2 2 6 0 - 4 4 3 11 1 - 7

Since the matrix is non-zero, its rank is at least equal to one.

The 2nd order minor - 1 1 2 2 = (- 1) × 2 - 1 × 2 = 4 is non-zero. This implies that the rank of the matrix A is at least two.

We sort through the minors of the 3rd order: C 3 3 × C 5 3 \u003d 1 5 ! 3! (5 - 3) ! = 10 pieces.

1 1 - 1 2 2 6 4 3 11 = (- 1) × 2 × 11 + 1 × 6 × 4 + (- 1) × 2 × 3 - (- 1) × 2 × 4 - 1 × 2 × 11 - (-1) × 6 × 3 = 0

1 - 1 - 2 2 6 0 4 11 1 = (- 1) × 6 × 1 + (- 1) × 0 × 4 + (- 2) × 2 × 11 - (- 2) × 6 × 4 - (- 1) × 2 × 1 - (- 1) × 0 × 11 = 0

1 1 - 2 2 2 0 4 3 1 = (- 1) × 2 × 1 + 1 × 0 × 4 + (- 2) × 2 × 3 - (- 2) × 2 × 4 - 1 × 2 × 1 - (-1) × 0 × 3 = 0

1 - 1 0 2 6 - 4 4 11 - 7 = (- 1) × 6 × (- 7) + (- 1) × (- 4) × 4 + 0 × 2 × 11 - 0 × 6 × 4 - ( - 1) × 2 × (- 7) - (- 1) × (- 4) × 11 = 0

1 - 1 0 2 6 - 4 3 11 - 7 = 1 × 6 × (- 7) + (- 1) × (- 4) × 3 + 0 × 2 × 11 - 0 × 6 × 3 - (- 1) × 2 × (- 7) - 1 × (- 4) × 11 = 0

1 - 2 0 2 0 - 4 3 1 - 7 = 1 × 0 × (- 7) + (- 2) × (- 4) × 3 + 0 × 2 × 1 - 0 × 0 × 3 - (- 2) × 2 × (- 7) - 1 × (- 4) × 1 = 0

1 - 2 0 6 0 - 4 11 1 - 7 = (- 1) × 0 × (- 7) + (- 2) × (- 4) × 11 + 0 × 6 × 1 - 0 × 0 × 11 - ( - 2) × 6 × (- 7) - (- 1) × (- 4) × 1 = 0

The 3rd order minors are zero, so the rank of the matrix is two.

Answer : Rank (A) = 2.

Finding the rank of a matrix by the method of fringing minors

Definition 3

Fringing Minor Method - a method that allows you to get a result with less computational work.

Fringing minor - minor M o k (k + 1) -th order of the matrix A, which borders the minor M of order k of the matrix A, if the matrix that corresponds to the minor M o k "contains" the matrix that corresponds to the minor M.

Simply put, the matrix corresponding to the bordered minor M is obtained from the matrix corresponding to the bordering minor M o k by deleting the elements of one row and one column.

Example 3

Find the rank of a matrix:

A = 1 2 0 - 1 3 - 2 0 3 7 1 3 4 - 2 1 1 0 0 3 6 5

To find the rank, we take the 2nd order minor M = 2 - 1 4 1

We write down all bordering minors:

1 2 - 1 - 2 0 7 3 4 1 , 2 0 - 1 0 3 7 4 - 2 1 , 2 - 1 3 0 7 1 4 1 1 , 1 2 - 1 3 4 1 0 0 6 , 2 0 - 1 4 - 2 1 0 3 6 , 2 - 1 3 4 1 1 0 6 5 .

To substantiate the method of bordering minors, we present a theorem whose formulation does not require a proof base.

Theorem 1

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.

Action algorithm :

To find the rank of a matrix, it is not necessary to go through all the minors, just look at the borders.

If the bordering minors are equal to zero, then the rank of the matrix is zero. If there exists at least one minor that is not equal to zero, then we consider bordering minors.

If they are all zero, then Rank(A) is two. If there is at least one nonzero bordering minor, then we proceed to consider its bordering minors. And so on, in a similar way.

Example 4

Find the rank of a matrix by the method of fringing minors

A = 2 1 0 - 1 3 4 2 1 0 - 1 2 1 1 1 - 4 0 0 2 4 - 14

How to decide?

Since the element a 11 of the matrix A is not equal to zero, then we take the minor of the 1st order. Let's start looking for a bordering minor other than zero:

2 1 4 2 = 2 x 2 - 1 x 4 = 0 2 0 4 1 = 2 x 1 - 0 x 4 = 2

We have found a bordering minor of the 2nd order which is not equal to zero 2 0 4 1 .

Let's enumerate the bordering minors - (there are (4 - 2) × (5 - 2) = 6 pieces).

2 1 0 4 2 1 2 1 1 = 0 ; 2 0 - 1 4 1 0 2 1 1 = 0 ; 2 0 3 4 1 - 1 2 1 - 4 = 0 ; 2 1 0 4 2 1 0 0 2 = 0 ; 2 0 - 1 4 1 0 0 2 4 = 0 ; 2 0 3 4 1 - 1 0 2 - 14 = 0

Answer : Rank(A) = 2.

Finding the rank of a matrix by the Gauss method (using elementary transformations)

Recall what elementary transformations are.

Elementary transformations:

by rearranging the rows (columns) of the matrix;
by multiplying all elements of any row (column) of the matrix by an arbitrary non-zero number k;

by adding to the elements of any row (column) elements that correspond to another row (column) of the matrix, which are multiplied by an arbitrary number k.

Definition 5

Finding the rank of a matrix using the Gauss method - a method based on the theory of matrix equivalence: 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 definition of the matrix:

in the case of a permutation of the rows or columns of a matrix, its determinant changes sign. If it is equal to zero, then when permuting rows or columns it remains equal to zero;
in the case of multiplying all elements of any row (column) of the matrix by an arbitrary number k, which is not equal to zero, the determinant of the resulting matrix is equal to the determinant of the original matrix, which is multiplied by k;

in the case of adding to the elements of a certain row or column of the matrix the corresponding elements of another row or column, which are multiplied by the number k, does not change its determinant.

The essence of the method of elementary transformations : reduce the matrix, whose rank is to be found, to a trapezoidal one using elementary transformations.

For what?

The rank of matrices of this kind is quite easy to find. It is equal to the number of rows that have at least one non-null element. And since the rank does not change during elementary transformations, this will be the rank of the matrix.

Let's illustrate this process:

for rectangular matrices A of order p by n, the number of rows of which is greater than the number of columns:

A ~ 1 b 12 b 13 ⋯ b 1 n - 1 b 1 n 0 1 b 23 ⋯ b 2 n - 2 b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bn - 1 n 0 0 0 ⋯ 0 1 0 0 0 ⋯ 0 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 , R ank (A) = n

A ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0 , R ank (A) = k

for rectangular matrices A of order p by n, the number of rows of which is less than the number of columns:

A ~ 1 b 12 b 13 ⋯ b 1 pb 1 p + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 pb 2 p + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bpp + 1 ⋯ bpn , R ank (A) = p

A ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0

for square matrices A of order n by n:

A ~ 1 b 12 b 13 ⋯ b 1 n - 1 b 1 n 0 1 b 23 ⋯ b 2 n - 1 b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bn - 1 n 0 0 0 ⋯ 0 1 , R ank (A) = n

A ~ 1 b 12 b 13 ⋯ b 1 kb 1 k + 1 ⋯ b 1 n 0 1 b 23 ⋯ b 2 kb 2 k + 1 ⋯ b 2 n ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 1 bkk + 1 ⋯ bkn 0 0 0 ⋯ 0 0 ⋯ 0 ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ 0 0 0 ⋯ 0 0 ⋯ 0 , R ank (A) = k , k< n

Example 5

Find the rank of matrix A using elementary transformations:

A = 2 1 - 2 6 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11

How to decide?

Since the element a 11 is non-zero, it is necessary to multiply the elements of the first row of the matrix A by 1 a 11 \u003d 1 2:

A = 2 1 - 2 6 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11 ~

We add to the elements of the 2nd row the corresponding elements of the 1st row, which are multiplied by (-3). To the elements of the 3rd row we add the elements of the 1st row, which are multiplied by (-1):

~ A (1) \u003d 1 1 2 - 1 3 3 0 0 - 1 1 - 1 2 - 7 5 - 2 4 - 15 7 2 - 4 11 ~ A (2) \u003d \u003d 1 1 2 - 1 3 3 + 1 (- 3) 0 + 1 2 (- 3) 0 + (- 1) (- 3) - 1 + 3 (- 3) 1 + 1 (- 3) - 1 + 1 2 (- 3) 2 + (- 1) (- 1) - 7 + 3 (- 1) 5 + 1 (- 5) - 2 + 1 2 (- 5) 4 + (- 1) (- 5) - 15 + 3 (- 5) 7 + 1 (- 7) 2 + 1 2 (- 7) - 4 + (- 1) (- 7) 11 + 3 (- 7) =

1 1 2 - 1 3 0 - 3 2 3 - 10 0 - 3 2 3 - 10 0 - 9 2 9 - 30 0 - 3 2 3 - 10

The element a 22 (2) is non-zero, so we multiply the elements of the 2nd row of the matrix A by A (2) by a 1 a 22 (2) = - 2 3:

A (3) \u003d 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 3 - 10 0 - 9 2 9 - 30 0 - 3 2 3 - 10 ~ A (4) \u003d 1 1 2 - 1 3 0 1 - 2 20 3 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 0 - 9 2 + 1 9 2 9 + (- 2) 9 2 - 30 + 20 3 × 9 2 0 - 3 2 + 1 3 2 3 + (- 2) 3 2 - 10 + 20 3 × 3 2 = = 1 1 2 - 1 3 0 1 - 2 20 3 0 0 0 0 0 0 0 0 0 0 0 0

To the elements of the 3rd row of the resulting matrix, we add the corresponding elements of the 2nd row, which are multiplied by 3 2 ;
to the elements of the 4th row - the elements of the 2nd row, which are multiplied by 9 2 ;
to the elements of the 5th row - the elements of the 2nd row, which are multiplied by 3 2 .

All row elements are zero. Thus, with the help of elementary transformations, we have reduced the matrix to a trapezoidal form, from which it can be seen that R a n k (A (4)) = 2 . It follows that the rank of the original matrix is also equal to two.

Comment

If you carry out elementary transformations, then approximate values are not allowed!

If you notice a mistake in the text, please highlight it and press Ctrl+Enter

Matrix rank is the largest order of its non-zero minors. The rank of a matrix is denoted by or .

If all order minors of a given matrix are zero, then all higher order minors of this matrix are also zero. This follows from the definition of the determinant. This implies an algorithm for finding the rank of a matrix.

If all first-order minors (elements of the matrix ) are equal to zero, then . If at least one of the first-order minors is different from zero, and all second-order minors are equal to zero, then . Moreover, it is enough to look through only those minors of the second order, which border the non-zero minor of the first order. If there is a second-order minor other than zero, one investigates the third-order minors surrounding the non-zero second-order minor. This continues until one of two cases is reached: either all the minors of order , bordering the non-zero minor of the -th order are equal to zero, or there are no such minors. Then .

Example 10 Calculate the rank of the matrix .

The first-order minor (element ) is different from zero. The minor that surrounds it is also non-zero.

All these minors are equal to zero, so .

The above algorithm for finding the rank of a matrix is not always convenient, since it involves the calculation of a large number of determinants. When calculating the rank of a matrix, it is most convenient to use elementary transformations, with the help of which the matrix is reduced to such a simple form that it is obvious what its rank is.

Elementary matrix transformations called the following transformations:

Ø multiplication of any row (column) of the matrix by a non-zero number;

Ø addition to one row (column) of another row (column), multiplied by an arbitrary number.

Half Jordan matrix row transformation:

with a resolving element, the following set of transformations with matrix rows is called:

Ø add u multiplied by a number to the first line, etc.;

Ø add u multiplied by the number to the last line.

Semi-Jordan transformation of matrix columns with a resolving element is called the following set of transformations with matrix columns:

Ø to the first column add th, multiplied by a number, etc .;

Ø to the last column add th, multiplied by the number.

After performing these transformations, the resulting matrix is:

Semi-Jordan transformation of rows or columns of a square matrix does not change its determinant.

Elementary transformations of a matrix do not change its rank. Let's show an example how to calculate the rank of a matrix using elementary transformations. rows (columns) are linearly dependent.

Let some matrix be given:

Select in this matrix arbitrary lines and arbitrary columns
. Then the determinant th order, composed of matrix elements
located at the intersection of selected rows and columns is called a minor -th order matrix
.

Definition 1.13. Matrix rank
is the largest order of the non-zero minor of this matrix.

To calculate the rank of a matrix, one should consider all its minors of the smallest order and, if at least one of them is nonzero, proceed to the consideration of minors of the highest order. This approach to determining the rank of a matrix is called the bordering method (or the bordering minors method).

Task 1.4. By the method of bordering minors, determine the rank of a matrix
.

Consider first-order bordering, for example,
. Then we turn to the consideration of some bordering of the second order.

For instance,
.

Finally, let's analyze the bordering of the third order.

So the highest order of a non-zero minor is 2, hence
.

When solving Problem 1.4, one can notice that the series of bordering minors of the second order are nonzero. In this regard, the following notion takes place.

Definition 1.14. The basis minor of a matrix is any non-zero minor whose order is equal to the rank of the matrix.

Theorem 1.2.(Basic minor theorem). Basic rows (basic columns) are linearly independent.

Note that the rows (columns) of a matrix are linearly dependent if and only if at least one of them can be represented as a linear combination of the others.

Theorem 1.3. The number of linearly independent matrix rows is equal to the number of linearly independent matrix columns and is equal to the rank of the matrix.

Theorem 1.4.(Necessary and sufficient condition for the determinant to be equal to zero). In order for the determinant -th order is equal to zero, it is necessary and sufficient that its rows (columns) be linearly dependent.

Calculating the rank of a matrix based on its definition is too cumbersome. This becomes especially important for high-order matrices. In this regard, in practice, the rank of a matrix is calculated based on the application of Theorems 10.2 - 10.4, as well as the use of the concepts of matrix equivalence and elementary transformations.

Definition 1.15. Two matrices
and are called equivalent if their ranks are equal, i.e.
.

If matrices
and are equivalent, then note
 .

Theorem 1.5. The rank of a matrix does not change from elementary transformations.

We will call elementary transformations of the matrix
any of the following actions on the matrix:

Replacing rows with columns and columns with corresponding rows;

Permutation of matrix rows;

Crossing out a line, all elements of which are equal to zero;

Multiplying any string by a non-zero number;

Adding to the elements of one row the corresponding elements of another row multiplied by the same number
.

Corollary of Theorem 1.5. If the matrix
obtained from the matrix using a finite number of elementary transformations, then the matrices
and are equivalent.

When calculating the rank of a matrix, it should be reduced to a trapezoidal form using a finite number of elementary transformations.

Definition 1.16. We will call trapezoidal such a form of representation of a matrix when in the bordering minor of the largest order other than zero, all elements below the diagonal ones vanish. For instance:

Here
, matrix elements
turn to zero. Then the form of representation of such a matrix will be trapezoidal.

As a rule, matrices are reduced to a trapezoidal shape using the Gaussian algorithm. The idea of the Gaussian algorithm is that, by multiplying the elements of the first row of the matrix by the corresponding factors, they achieve that all elements of the first column located below the element
, would turn to zero. Then, multiplying the elements of the second column by the corresponding multipliers, we achieve that all elements of the second column located below the element
, would turn to zero. Further proceed similarly.

Task 1.5. Determine the rank of a matrix by reducing it to a trapezoidal form.

For the convenience of applying the Gaussian algorithm, you can swap the first and third rows.








.

Obviously here
. However, to bring the result to a more elegant form, further transformations over the columns can be continued.










.

The number r is called the rank of the matrix A if:
1) the matrix A contains a non-zero minor of order r;
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 order of a non-zero minor.
Designations: rangA , r A or r .
It follows from the definition that r is a positive integer. For a null matrix, the rank is considered to be zero.

Service assignment. The online calculator is designed to find matrix rank. The solution is saved in Word and Excel format. see solution example.

Instruction. Select the dimension of the matrix, click Next.

Definition . Let a matrix of rank r be given. Any matrix minor other than zero and of order r is called basic, and the rows and columns of its components are called basic rows and columns.
According to this definition, the matrix A can have several basis minors.

The rank of the identity matrix E is n (number of rows).

Example 1 . Given two matrices, and their minors , . Which of them can be taken as the basis?
Solution. The minor M 1 =0, so it cannot be a basis for any of the matrices. Minor M 2 =-9≠0 and has order 2, so it can be taken as the basis matrices of 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 3, that is, M 2 is not a basis for the matrix A . Note that the matrix A has a unique basis minor equal to the determinant of the matrix A .

Theorem (on the basis minor). Any row (column) of a matrix is a linear combination of its basic rows (columns).
Consequences from the theorem.

Any (r+1) columns (rows) of a matrix of rank r are 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 another row (column) multiplied by any number other than zero is added to a row (column) of a matrix, then the rank of the matrix will not change.
If you cross out a row (column) in the 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 rank of a matrix, we will look for a minor of the highest order that is different from zero. First, we transform the matrix to a simpler form. To do this, multiply the first row of the matrix by (-2) and add to the second, then multiply it by (-1) and add to the third.