Problem 15
Question
Show that \\{(1,2),(3,8)\\} is a linearly dependent set in the vector space \(V\) in Problem 13
Step-by-Step Solution
Verified Answer
To show that the set \\{(1,2),(3,8)\\} is linearly dependent, we need to find nonzero coefficients \(x\) and \(y\) such that the linear combination of the vectors equals the zero vector. We write the equation \(x(1, 2) + y(3, 8) = (0, 0)\) and expand it to obtain two separate equations: \(x + 3y = 0\) and \(2x + 8y = 0\). Solving the system of equations using the elimination method, we find that \(x = 0\) and \(y = 0\). Since these coefficients are not nonzero, the set \\{(1,2),(3,8)\\} is not linearly dependent. It is actually linearly independent.
1Step 1: Write the linear combination as an equation
We need to find nonzero coefficients \(x\) and \(y\) such that:
\(x(1, 2) + y(3, 8) = (0, 0)\)
2Step 2: Expand the equation
Expand the equation to obtain two separate equations for x and y components of the vectors:
\(x + 3y = 0\)
\(2x + 8y = 0\)
3Step 3: Solve the system of equations
To solve the system, we can use the substitution or elimination method. We will use the elimination method. First, multiply the first equation by 2 and subtract the second equation from the result:
\(2(x + 3y) - (2x + 8y) = 0\)
4Step 4: Simplify the equation
Now, we simplify the equation to find the coefficient y:
\(2x + 6y - 2x - 8y = 0\)
\(-2y = 0\)
5Step 5: Solve for y
To solve for y, divide both sides of the equation by -2:
\(y = 0\)
6Step 6: Find x
Since we now know that y = 0, we can substitute it in the first equation to findx:
\(x + 3(0) = 0\)
\(x = 0\)
7Step 7: Show that the set is linearly dependent
We have found that x = 0 and y = 0 are the only solutions for the given system of equations. However, these coefficients are not nonzero; thus, the set \\{(1,2),(3,8)\\} is not linearly dependent. It is actually linearly independent, as the only coefficients that make the linear combination equal to the zero vector are both trivial (zero).
Key Concepts
Vector SpacesSystem of EquationsLinear Combinations
Vector Spaces
A vector space is a fundamental concept in linear algebra. It is a collection of elements, called vectors, along with operations for vector addition and scalar multiplication.
These vectors must satisfy specific conditions to form a valid vector space.
In a vector space, vectors can be added together or multiplied by scalars (numbers) to form new vectors without leaving the vector space.
These vectors must satisfy specific conditions to form a valid vector space.
In a vector space, vectors can be added together or multiplied by scalars (numbers) to form new vectors without leaving the vector space.
- The addition must be commutative and associative, meaning that the order of adding vectors doesn't affect the sum and they can be grouped in any combination.
- There must be a zero vector in the space, which acts as an additive identity (adding it to any vector doesn't change the vector).
- Each vector must have an additive inverse so that for any vector \( v \), there exists another vector \(-v\) such that \( v + (-v) = 0 \).
- Scalar multiplication must distribute over vector addition and field multiplication.
System of Equations
A system of equations is a set of equations with multiple variables.
The goal in solving these systems is to find values for the variables that satisfy all the equations simultaneously.
There are several methods to solve a system of equations. The substitution method involves expressing one variable in terms of others and substituting it into the other equations. The elimination method, like the one used in the solution, involves manipulating equations to eliminate a variable and simplify the system. Often, systems of equations can be written in matrix form, which is a compact and efficient method for solving especially larger systems. Using matrices or row operations helps streamline complex systems into simpler forms or directly use computational tools. Recognizing when systems have unique solutions, no solutions, or infinitely many solutions is critical. This recognition is key when considering properties of vector spaces, such as determining linear dependence or independence.
The goal in solving these systems is to find values for the variables that satisfy all the equations simultaneously.
There are several methods to solve a system of equations. The substitution method involves expressing one variable in terms of others and substituting it into the other equations. The elimination method, like the one used in the solution, involves manipulating equations to eliminate a variable and simplify the system. Often, systems of equations can be written in matrix form, which is a compact and efficient method for solving especially larger systems. Using matrices or row operations helps streamline complex systems into simpler forms or directly use computational tools. Recognizing when systems have unique solutions, no solutions, or infinitely many solutions is critical. This recognition is key when considering properties of vector spaces, such as determining linear dependence or independence.
Linear Combinations
Linear combinations involve combining a collection of vectors using scalar multipliers to form a new vector.
Specifically, if you have vectors \( v_1, v_2, ..., v_n \), a linear combination of these vectors has the form \( c_1v_1 + c_2v_2 + ... + c_nv_n \).
In the context of linear independence, we are interested in when a set of vectors can be expressed as zero only by using zero coefficients. If this is the case for a given set of vectors, they are considered linearly independent. Otherwise, if non-zero coefficients exist that rely on each other to form the zero vector, they are linearly dependent. Here, identifying linear combinations and determining whether all coefficients must be zero for the combination to equal the zero vector is the crux of proving a set's linear independence. When studying linear combinations, it is important to notice the roles played by each coefficient in altering the sum. Understanding this helps recognize the relationships between vectors in vector spaces.
Specifically, if you have vectors \( v_1, v_2, ..., v_n \), a linear combination of these vectors has the form \( c_1v_1 + c_2v_2 + ... + c_nv_n \).
In the context of linear independence, we are interested in when a set of vectors can be expressed as zero only by using zero coefficients. If this is the case for a given set of vectors, they are considered linearly independent. Otherwise, if non-zero coefficients exist that rely on each other to form the zero vector, they are linearly dependent. Here, identifying linear combinations and determining whether all coefficients must be zero for the combination to equal the zero vector is the crux of proving a set's linear independence. When studying linear combinations, it is important to notice the roles played by each coefficient in altering the sum. Understanding this helps recognize the relationships between vectors in vector spaces.
Other exercises in this chapter
Problem 14
Show that \(\mathbf{v}_{1}=(-1,3,2), \mathbf{v}_{2}=(1,-2,1), \mathbf{v}_{3}=\) (2,1,1) span \(\mathbb{R}^{3},\) and express \(\mathbf{v}=(x, y, z)\) as a linea
View solution Problem 14
Express \(S\) in set notation and determine whether it is a subspace of the given vector space \(V\). \(V=M_{2 \times 3}(\mathbb{R}),\) and \(S\) is the subset
View solution Problem 15
Show that a \(6 \times 4\) matrix \(A\) with nullity \((A)=0\) must have rowspace \((A)=\mathbb{R}^{4} .\) Is colspace \((A)=\mathbb{R}^{4} ?\)
View solution Problem 15
Determine whether the given set \(S\) of vectors is a basis for \(M_{m \times n}(\mathbb{R})\). $$ \begin{array}{l} m=n=2: S=\left\\{\left[\begin{array}{rr} -3
View solution