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The rank of a matrix in checking whether vectors form a basis of a linear space

Order of Matrices Lecture 2

Topic: Checking whether vectors form a basis in a linear space

Summary

In this article, I’ll show you how to use the rank of a matrix to check whether vectors form a basis of a linear space.

When do vectors form a basis in linear space?

We call several vectors a basis of a linear space when two conditions are met:

  1. These vectors are linearly independent (this is where the order of the matrix is ​​used)
  2. Any other vector in the same space can be represented as a linear combination of these vectors

So to check if several vectors form a basis, you have to show that they satisfy these two conditions.

Calculation of linear independence of vectors

Here the matter is simple. Vectors are arranged by rows (or columns) in a matrix. We count the row of the matrix. The order of a matrix is ​​the maximum number of independent vectors that compose it. So if the rank of a matrix is ​​equal to the number of vectors in it, then they are linearly independent. If it does not come out equal – they are linearly dependent (and do not form a basis in a given space).

Proving that every vector can be expressed as a linear combination of linearly independent vectors

Here we take some arbitrary vector in a given space and show that there are (or not) constants that multiplied by the base vectors added together will be equal to this arbitrary vector. How this is done – I will show further in the example.

Example (task)

For example, we can do the following task:

Excercise 1

Check that the vectors: \vec{x_1} =[1,0,3], \vec{x_2}=[0,2,2], \vec{x_3} =[1,-2,0] form a basis in linear space {\R}^{3}, and if so, represent the vector \vec{a}=[10,2,4] as their linear combination.

We start by checking the linear independence of these vectors. We put them in a matrix and count its rank :

Order of matrices in the study of linear independence of vectorsWe count, count, count (how to do it exactly is shown, for example, in my Matrix Course) and we have the result:

The calculated rank of the matrixIf the rank of the matrix is ​​3, and the vectors are also three, then they are linearly independent . The first condition for the vectors to form a basis in a linear space {\R}^3 is fulfilled.

In the second condition we have to show that any vector belonging to the space {\R}^3 (i.e. three-dimensional) can be represented as the sum of vectors forming – supposedly – a base multiplied by constants. You can write this condition as:

{\forall}_{\vec{x}}{\exists}_{{\alpha_1},{\alpha_2},{\alpha_3}}\vec{x}={\alpha_1}\vec{x_1}+{\alpha_2}\vec{x_2}+{\alpha_3}\vec{x_3}

Switching to the notation with coordinates of vectors and their column form (more convenient), this condition will look like this:

Linear combination of vectorsMultiplying by constants we get:

Linear combination after multiplication by constantsAnd after adding the matrices on the right:

Linear combination after matrix additionThe above equation will give us a system of equations:

System of equations for a linear combination of vectorsNote that this system ALWAYS has a solution, no matter how much x,y,z are equal, because the main determinant of the system (from Cramer’s formulas) is different from zero:

Principal determinant of a system of linear equationsSo for any x,y,z there will be one {\alpha_1} ,{\alpha_2} ,{\alpha_3} that the system will have a solution. So any vector in a linear space {\R}^3 can be represented as a linear combination of vectors \vec{x_1} ,\vec{x_2} ,\vec{x_3}.

So we have the answer. Vectors \vec{x_1} ,\vec{x_2} ,\vec{x_3} form a basis in linear space {\R}^3.

Let’s recall the command:

Check that the vectors: \vec{x_1} =[1,0,3], \vec{x_2}=[0,2,2], \vec{x_3}=[1,-2,0] form a basis in linear space {\R}^3, and if so, represent the vector \vec{a}=[10,2,4] as their linear combination.

So our mission is not over yet, we still need a vector \vec{a} =[10,2,4] represent as a linear combination of vectors \vec{x_1} =[1,0,3], \vec{x_2} =[0,2,2], \vec{x_3} =[1,-2,0] that is, find such a constant \alpha_1, \alpha_2, \alpha_3, That:

Vector from the problem as a linear combination of vectorsAfter multiplying by vector constants on the right:

Linear combination after multiplication by constantsAfter adding the vectors on the right:

Linear combination of vectors after vector additionWhat can be written as a system of equations:

System of equations for a linear combinationWe solve this system – for example – by Cramer’s formulas and we get the result:

Solving the system of equations - calculated constantsSo our vector represented as a linear combination of three vectors can be written as:

Answer - a linear combination of vectorsAnd this is our answer.

Let’s do an example where the vectors will NOT be linearly independent, as shown by the order of the matrix.

Exercise 2

Check that the vectors: \vec{x_1} =[4,-1,4], \vec{x_2} =[3,2,5], \vec{x_3} =[8,-2,8] form a basis in linear space {\R}^3.
To check the linear independence of the vectors, we calculate the rank of the appropriate matrix:

Matrix order (linearly dependent vectors)We count, count, count and we have the result: 2. So only 2 of these 3 are linearly independent. So these 3 vectors as a whole are not linearly independent. Thus, they do not form the basis of space {\R}^3 . Case closed, we can write a nice, finish answer.

Click to remember what the rank of a matrix is ​​(previous Lecture)< —

Click to see how the rank of a matrix can be applied to systems of linear equations (next Lecture) –>

Click to return to the page with Lectures to the Matrix

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