Linear Combinations and Spanning Sets

Author: Bindeshwar Singh Kushwaha

Institute: PostNetwork Academy

Introduction:

In this post, we will explore the concepts of Linear Combinations and Spanning Sets in the context of vector spaces in Linear Algebra.

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Linear Combinations and Spanning Sets

Let V be a vector space over a field K.

A vector v in V is a linear combination of vectors u₁, u₂, …, u_m in V if there exist scalars a₁, a₂, …, a_m in K such that:

\[
v = a_1 u_1 + a_2 u_2 + \cdots + a_m u_m
\]

Alternatively, v is a linear combination of u₁, u₂, …, u_m if there is a solution to:

\[
v = x_1 u_1 + x_2 u_2 + \cdots + x_m u_m
\]

where x₁, x₂, …, x_m are unknown scalars.

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Example: Linear Combination in R³

Suppose we want to express v = (3, 7, -4) in R³ as a linear combination of the vectors:

• • u₁ = (1, 2, 3)
  • u₂ = (2, 3, 7)
  • u₃ = (3, 5, 6)
  • Seek scalars x, y, z such that:

v = x u₁ + y u₂ + z u₃

Matrix Equation Form:

\[
\begin{bmatrix}
3 \\
7 \\
-4
\end{bmatrix}
=
x
\begin{bmatrix}
1 \\
2 \\
3
\end{bmatrix}
+
y
\begin{bmatrix}
2 \\
3 \\
7
\end{bmatrix}
+
z
\begin{bmatrix}
3 \\
5 \\
6
\end{bmatrix}
\]

Or the system of equations:

\[
\begin{cases}
x + 2y + 3z = 3 \\
2x + 3y + 5z = 7 \\
3x + 7y + 6z = -4
\end{cases}
\]

Row Reduction and Solution:

After reducing the system to echelon form, we get:

\[
\begin{cases}
x + 2y + 3z = 3 \\
-y – z = 1 \\
y – 3z = -13
\end{cases}
\]

Back-substitution yields the solution:

z = 3, y = -4, x = 2

Thus, v = 2u₁ – 4u₂ + 3u₃

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Example: Linear Combination in P(t)

Suppose we want to express the polynomial v = 3t² + 5t – 5 as a linear combination of:

• • p₁ = t² + 2t + 1
  • p₂ = 2t² + 5t + 4
  • p₃ = t² + 3t + 6
  • Seek scalars x, y, z such that:

v = x p₁ + y p₂ + z p₃

Expanding the right-hand side:

\[
x(t^2 + 2t + 1) + y(2t^2 + 5t + 4) + z(t^2 + 3t + 6)
\]

Combine like terms:

\[
= (x + 2y + z)t^2 + (2x + 5y + 3z)t + (x + 4y + 6z)
\]

Equating coefficients with 3t² + 5t – 5 gives the system:

\[
\begin{cases}
x + 2y + z = 3 \\
2x + 5y + 3z = 5 \\
x + 4y + 6z = -5
\end{cases}
\]

Row Reduction and Solution:

After reducing the system:

\[
\begin{cases}
x + 2y + z = 3 \\
y + z = -1 \\
3z = -6
\end{cases}
\]

Solving by back-substitution:

z = -2, y = 1, x = 3

Thus, v = 3p₁ + p₂ – 2p₃

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Spanning Sets

Let V be a vector space over a field K. Vectors u₁, u₂, …, u_m in V are said to span V if every vector in V is a linear combination of them:

\[
v = a_1 u_1 + a_2 u_2 + \cdots + a_m u_m
\]

Remark 1: If u₁, …, u_m span V, then so does w, u₁, …, u_m for any w in V.

Remark 2: If u_k is a linear combination of the others, removing it still spans V.

Remark 3: If one u_k = 0, then removing it still spans V.

PDF

LinearCombinationsSpanningSets

Video

 

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