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A group of vectors is called linearly independent when none of the vectors can be made by adding or scaling the others. In simple terms, every vector in the set gives new information and doesn’t repeat or depend on the others.
If you list several vectors, and you cannot write any one of them as a combination (like multiplying and adding) of the ones before it, then that group is linearly independent. This idea is very important in mathematics, especially when working with vector spaces.
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In the physical world, some quantities like temperature or mass have only size or amount. These are called scalars, and we usually represent them using regular numbers.
But other quantities, like force or velocity, have both size and direction. These are called vectors. You can think of a vector as an arrow starting from a point, with its length showing how big it is, and its direction showing which way it’s going.
A vector space V is a collection of objects with a (vector) addition and scalar multiplication defined that closed under both operations and which in addition satisfies the following axioms:
(i) ({\alpha} + {\beta})x = {\alpha}x + {\beta}x for all x {\in} V and {\alpha}, {\beta} {\in} F
(ii) {\alpha}({\beta}x)=({\alpha}{\beta})x
(iii) x + y = y + x for all x, y {\in} V
(iv) x + (y + z)=(x + y) + z for all x, y, z {\in} V
(v) {\alpha}(x + y) = {\alpha}x + {\alpha}y
(vi) ∃O {\in} V z 0 + x = x; 0 is usually called the origin
(vii) 0x = 0
(viii) ex = x where e is the multiplicative unit in F
A linearly dependent vector can be defined as:
A group of vectors is called linearly dependent when at least one of the vectors can be written using the others in the group. In other words, if one vector is a combination of the rest (like added or multiplied by numbers), then they are dependent.
Some facts about linear dependence are:
The determinant of the linearly dependent matrix is zero. Mathematically we can write it as:
\begin{vmatrix}\overrightarrow{u_1} &\overrightarrow {u_2} &\overrightarrow {u_3} \overrightarrow {v_1} & \overrightarrow {v_2} & \overrightarrow {v_3} \overrightarrow {w_1} & \overrightarrow {w_2} & \overrightarrow {w_3} \end{vmatrix} = 0
Let us summarize the properties of linearly dependent vectors.
The linear independence of a set of vectors can be determined by calculating the determinant of a matrix with columns composed of the vectors in the set. If the determinant is equal to zero, then the set of vectors is linearly dependent. If the determinant is non-zero, then the set of vectors is linearly independent.
Let A = { v_1, v_2, …, v_k } be a collection of vectors from R_n . If k > 2 and at least one of the vectors in A can be written as a linear combination of the others, then A is said to be linearly dependent.
A set of vectors { v_1, v_2, …, v_k } is linearly independent if the vector equation
x_1v_1+x_2v_2+…+x_kv_k=0
has only the trivial solution x_1=x_2=…x_k=0. The set {v_1,v_2,…,v_k} is linearly dependent otherwise.
In other words, {v_1,v_2,…,v_k} is linearly dependent if there exist numbers x_1, x_2, …, x_k, not all equal to zero, such that
x_1v_1+x_2v_2+…+x_kv_k=0
This is called a linear dependence relation or equation of linear dependence.
If every element of V can be expressed in a unique way as a finite linear combination of elements of B, then the set B of vectors in the vector space V is referred to as a basis in mathematics. The components or coordinates of the vector with respect to B are what is known as the coefficients of this linear combination.
Let V be a subspace of R_n for some n. A collection B = { v_1, v_2, …, v_k } of vectors from V is said to be a basis for V if B is linearly independent and spans V. If either one of these criteria is not satisfied, then the collection is not a basis for V. If a collection of vectors spans V, then it contains enough vectors so that every vector in V can be written as a linear combination of those in the collection.
The cardinality of a basis of a vector space V over its base field, or the number of vectors, determines the dimension of a vector space V in mathematics. To distinguish it from other types of dimensions, it is frequently referred to as the Hamel dimension (after Georg Hamel) or algebraic dimension.
Linear independence plays an important role in many areas of science, engineering, and technology. It helps us understand if a set of vectors or equations provide unique, useful information or if some of them are just repeating what others already show.
In machine learning, it is important to make sure that the input features (or variables) are not repeating the same patterns. Linearly independent features help avoid problems like multicollinearity, where some features overlap. This ensures that each feature contributes new, useful information for better predictions.
In computer graphics, linear independence helps in creating movements and effects by using transformations. For example, when rotating or scaling images, independent vectors help define directions and dimensions clearly, making the visuals more accurate and realistic.
Scientists and engineers apply linear independence to analyze systems of motions or forces. It serves to decompose complex situations into fewer parts in these disciplines, which simplifies designing buildings, bridges, machines, or even replicating real-world occurrences such as waves or earthquakes.
Linear independence can be used to see if a system of equations has exactly one solution. If all the equations are independent, then you can have one distinct solution. If they're not independent, then you could have no solution or many solutions.
In such regions as signal processing or image compression, employment of linearly independent vectors enhances the possibility of reduced storage and clean signals. This contributes to minimizing data size without sacrificing vital details — applicable in phones, media applications, and communication systems.
Here are some solved examples of Linearly Independent Vectors.
Example 1: Determine the values of k for the linearly dependent vectors {\overrightarrow} {u} = (3, k , -6), {\overrightarrow} {v} = (-2, 1, k + 3) and {\overrightarrow}{w} = (1, k + 2, 4).
Solution: We know that the vectors are linearly dependent if the determinant of the matrix is zero, meaning that the rank of the matrix is less than 3.
\begin{vmatrix} 3 & k & -6 -2 & 1 & k + 3 1 & k + 2 & 4 \end{vmatrix}= 0
Recall the formula for finding the determinant of a 3×3 matrix and use it to find the determinant of the above matrix:
= 3\cdot\begin{vmatrix} 1 & k + 3 \\ k + 2 & 4\end{vmatrix} – k \cdot \begin{vmatrix} -2 & k + 3\\ 1 & 4\end{vmatrix} + (-6)\begin{vmatrix} -2 & 1 \\1 & k + 2\end{vmatrix}=0
12 + k^2 + 3k + 12k + 24 – (-6 – 8k + 3k^2 + 15k + 18) = 0
k^2 – 4k – 12 = 0
k = -2
k = 6
Example 2: Determine whether the following vectors are linearly dependent or independent.
{\overrightarrow} {u} = (1, 3, 0) , {\overrightarrow} {v} = (2, 0, 1), {\overrightarrow} {w} = (3, 3, 1)
Solution: In the first step of solving this problem, we will multiply the vectors u, v and w with a, b and c respectively in such a way that a + b + c = 0
a (1, 3, 0) + b (2, 0, 1) + c (3, 3, 1) = (0, 0, 0)
Write the above equation as a system of equations like this:
a+ 3b = 0
2a + c = 0
3a + 3b + c = 0
Now, we will write the equation is matrix form to find the determinant:
\begin{vmatrix} 1 & 3 & 0 \\ 2 & 0 & 1 \\ 3 & 3 & 1 \end{vmatrix}
Recall the formula for finding the determinant of the 3 x 3 matrix and utilize it to compute the determinant of the matrix form of the above equation:
= 1\cdot\begin{vmatrix} 0 & 1 \\ 3 & 1\end{vmatrix} – 3\cdot\begin{vmatrix} 2 & 1 \\ 3 & 1\end{vmatrix} + 0\begin{vmatrix} 2 & 0 \\ 3 & 3\end {vmatrix}
= 0
The zero determinant shows that the vectors are linearly dependent.
Example 3: Determine the values of t for the linearly dependent vectors {\overrightarrow} {u} = (2, t , 4), {\overrightarrow} {v} = (1, 1, t + 1) and {\overrightarrow}{w} = (2, t + 3, 4).
Solution: If the determinant of the matrix is zero, then vectors are linearly dependent. It also means that the rank of the matrix is less than 3. Hence, write the vectors in matrix form and set the matrix equal to zero like this:
\begin{vmatrix} 2 & t & 4 \\ 1 & 1 & t + 1\\ 2 & t + 3 & 4 \end{vmatrix} = 0
Recall the formula of finding the determinant of a 3×3 matrix and use it to find the determinant of the above matrix:
= 2\cdot\begin {vmatrix} 1 & t + 1\\t + 3 & 4\end {vmatrix} – t\cdot \begin {vmatrix} 1 & t + 1\\ 2 & 4\end {vmatrix} + 4 \begin {vmatrix} 1 & 1 \\2 & t + 3\end {vmatrix} = 0
-6t + 6 = 0
-6t = -6
t = 1
Hence, for t = 1, the vectors will be linearly dependent vectors.
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