Linear Algebra For Computer Science: Determine Whether
Linear Algebra For Computer Science Cat Iia Determine Whetheru282
Linear Algebra For Computer Science Cat II a) Determine whether u =(2,8,2) is a linear combination of u1 =(2,-2,0), u2 =(3,0,3), and u3 =(-2,0,-1) a. or not (5 marks) b) Show that the set of vectors k j a 2 - = r , k j i a + - = r and k j i c + + = 2 r is linearly independent (6 marks) c) What do you understand by the phrase “u is a linear combination of u1, u2,u3,…un †(3 marks) d) Determine whether the vector is a linear combination of the vectors and (5 marks) e) Consider the set B= {(1,1,1), (0,2,2), (0,0,3)}, show that B i. Spans R marks) ii. is linearly independent (3 marks) _.unknown _.unknown _.unknown
Paper For Above instruction
Introduction
Linear algebra is fundamental in computer science, especially in areas involving data representation, transformation, and various computational algorithms. Central concepts such as linear combinations, linear independence, span, and basis are crucial in understanding the structure of vector spaces. This paper explores these concepts through specific problems involving vectors in \(\mathbb{R}^3\) and discusses their applications in computational contexts.
Problem a: Determining if a vector is a linear combination of other vectors
The first problem asks whether the vector \(\mathbf{u} = (2,8,2)\) can be expressed as a linear combination of the vectors \(\mathbf{u}_1 = (2,-2,0)\), \(\mathbf{u}_2 = (3,0,3)\), and \(\mathbf{u}_3 =(-2,0,-1)\). Formally, we need to find scalars \(a, b,\) and \(c\) such that:
\[
a \mathbf{u}_1 + b \mathbf{u}_2 + c \mathbf{u}_3 = \mathbf{u}
\]
which translates into the system of equations:
\[
2a + 3b - 2c = 2
\]
\[
-2a + 0b + 0c = 8
\]
\[
0a + 3b - c = 2
\]
From the second equation:
\[
-2a = 8 \quad \Rightarrow \quad a = -4
\]
Substituting \(a = -4\) into the first and third equations:
\[
2(-4) + 3b - 2c = 2 \quad \Rightarrow \quad -8 + 3b - 2c = 2
\]
\[
\Rightarrow 3b - 2c = 10
\]
and
\[
3b - c = 2
\]
Expressing \(c\) from the second:
\[
c = 3b - 2
\]
Substituting into the first:
\[
3b - 2(3b - 2) = 10
\]
\[
3b - 6b + 4 = 10
\]
\[
-3b = 6 \quad \Rightarrow \quad b = -2
\]
then
\[
c = 3(-2) - 2 = -6 - 2 = -8
\]
The scalars are:
\[
a = -4, \quad b = -2, \quad c = -8
\]
Since these satisfy the equations, \(\mathbf{u}\) is a linear combination of \(\mathbf{u}_1, \mathbf{u}_2,\) and \(\mathbf{u}_3\).
Problem b: Show that a set of vectors is linearly independent
Next, consider the vectors:
\[
\mathbf{v}_1 = (a, 2, -), \quad \mathbf{v}_2 = (i, a, +), \quad \mathbf{v}_3 = (c, +, +)
\]
(Note: Since the problem as stated contains symbols that are unclear, we interpret these as generic vectors \(k j a 2 - = r\), \(k j i a + - = r\), and \(k j i c + + = 2r\). For clarity, assume the vectors are:
\[
\mathbf{v}_1 = (a, 2, -), \quad \mathbf{v}_2 = (i, a, +), \quad \mathbf{v}_3 = (c, +, +)
\]
assuming '+' and '-' are placeholders for specific values or variables).
To verify linear independence, we set up the matrix with these vectors as columns and check if the determinant is non-zero:
\[
\det \begin{bmatrix}
a & i & c \\
2 & a & + \\
- & + & +
\end{bmatrix} \neq 0
\]
Given the ambiguity in the vectors, the general principle is, if the determinant of the matrix formed by the vectors is not zero, the set is linearly independent. Alternatively, one can perform row operations to verify this condition.
Assuming the vectors are linearly independent, it implies no vector in the set can be written as a linear combination of the others. This property is fundamental when establishing bases in vector spaces, as linearly independent sets span the space with minimal redundancy.
Problem c: Understanding linear combinations
A vector \(\mathbf{u}\) is called a linear combination of vectors \(\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n\) if there exist scalars \(c_1, c_2, \ldots, c_n\) such that:
\[
\mathbf{u} = c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + \cdots + c_n \mathbf{u}_n
\]
This concept is fundamental in linear algebra because it expresses how vectors in a space can be constructed from a set of basic vectors. In computational contexts, linear combinations are used in transformations, modeling, and solving systems of equations.
Problem d: Determining if a vector is a linear combination of given vectors
Without the specific vectors detailed in the original question, the general approach involves solving the linear system similar to Problem a. Given a target vector \(\mathbf{w}\) and a set \(\{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k\}\), we check whether there exist scalars \(x_1, x_2, \ldots, x_k\) such that:
\[
x_1 \mathbf{v}_1 + x_2 \mathbf{v}_2 + \cdots + x_k \mathbf{v}_k = \mathbf{w}
\]
This involves setting up and solving linear equations, typically via matrix methods such as Gaussian elimination.
Problem e: Set B and its span and independence
The set:
\[
B = \{(1,1,1), (0,2,2), (0,0,3)\}
\]
is examined for spanning \(\mathbb{R}^3\) and linear independence.
Using matrix columns:
\[
\begin{bmatrix}
1 & 0 & 0 \\
1 & 2 & 0 \\
1 & 2 & 3
\end{bmatrix}
\]
To check spanning, we verify whether the vectors can generate any vector in \(\mathbb{R}^3\). Since the matrix is of full rank (determinant \(\neq 0\)), the vectors span \(\mathbb{R}^3\). The determinant:
\[
\det = 1 \times \det \begin{bmatrix} 2 & 0 \\ 2 & 3 \end{bmatrix} = 1 \times (2 \times 3 - 0 \times 2) = 6 \neq 0
\]
Thus, \(B\) spans \(\mathbb{R}^3\).
For linear independence, since the determinant of the matrix of vectors is non-zero, the set is linearly independent. This indicates that these vectors form a basis for \(\mathbb{R}^3\).
Conclusion
Understanding linear combinations, independence, and spanning sets is essential in linear algebra applications within computer science, such as graphics, machine learning, and data analysis. The problems examined demonstrate core principles: determining dependencies among vectors, representing vectors as combinations of others, and establishing bases for vector spaces. These concepts underpin many algorithms and data structures in computational disciplines.
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