Proving Linear Independence Of Vectors I+2j+3k, 2i+j+3k, And I+j+k

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Hey guys! Today, we're diving into a classic problem in linear algebra: proving that a set of vectors is linearly independent. Specifically, we'll be tackling the set of vectors: $\vec{i}+2 \vec{j}+3 \vec{k}, 2 \vec{i}+\vec{j}+3 \vec{k}$, and $\vec{i}+\vec{j}+\vec{k}$. Linear independence is a fundamental concept in vector spaces, and mastering it is crucial for understanding higher-level topics. So, let's break it down step by step, making sure everyone understands the process. Grab your thinking caps, and let's get started!

Understanding Linear Independence

Before we jump into the specific problem, let's make sure we're all on the same page about what linear independence actually means. Simply put, a set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. Think of it like this: each vector in the set contributes a unique direction, and you can't reach any vector by just combining the others. On the flip side, if a set of vectors is linearly dependent, it means at least one vector can be written as a linear combination of the others. This implies redundancy in the set, as one vector's direction can be obtained from the others.

Why is this important? Linear independence is crucial in many areas of mathematics and physics. For example, in linear algebra, a basis for a vector space is a set of linearly independent vectors that span the entire space. This means that any vector in the space can be uniquely represented as a linear combination of the basis vectors. In physics, linearly independent vectors can represent forces, velocities, or other physical quantities that act in different directions. The ability to determine if a set of vectors is linearly independent is essential for solving various problems in these fields.

To formally define linear independence, consider a set of vectors {$\vec{v}_1, \vec{v}_2, ..., \vec{v}_n$}. These vectors are linearly independent if the only solution to the equation:

c1v⃗1+c2v⃗2+...+cnv⃗n=0⃗c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_n\vec{v}_n = \vec{0}

is the trivial solution, where all the scalars $c_1, c_2, ..., c_n$ are equal to zero. In other words, the only way to get the zero vector by combining these vectors is by multiplying each of them by zero. If there exists a non-trivial solution (i.e., at least one $c_i$ is non-zero), then the vectors are linearly dependent.

In the context of the given problem, we need to show that the vectors $\vec{i}+2 \vec{j}+3 \vec{k}, 2 \vec{i}+\vec{j}+3 \vec{k}$, and $\vec{i}+\vec{j}+\vec{k}$ are linearly independent. This means we need to demonstrate that the only solution to the equation:

c1(i⃗+2j⃗+3k⃗)+c2(2i⃗+j⃗+3k⃗)+c3(i⃗+j⃗+k⃗)=0⃗c_1(\vec{i}+2 \vec{j}+3 \vec{k}) + c_2(2 \vec{i}+\vec{j}+3 \vec{k}) + c_3(\vec{i}+\vec{j}+\vec{k}) = \vec{0}

is $c_1 = c_2 = c_3 = 0$. If we can prove this, we've successfully shown that the vectors are linearly independent. Let's move on to the methods we can use to do this.

Methods to Prove Linear Independence

Okay, so we know what linear independence is, but how do we prove it? There are a couple of common methods, and we'll use the most straightforward one for this problem: setting up a system of equations. Let's briefly touch on the other methods as well.

1. Setting up a System of Equations

This is the most common and often the most intuitive method. As we discussed earlier, to prove linear independence, we start with the equation:

c1v⃗1+c2v⃗2+...+cnv⃗n=0⃗c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_n\vec{v}_n = \vec{0}

where $\vec{v}_1, \vec{v}_2, ..., \vec{v}_n$ are the vectors in question and $c_1, c_2, ..., c_n$ are scalar coefficients. We then expand this equation and group the components corresponding to each basis vector (in this case, $\vec{i}, \vec{j}, \vec{k}$). This results in a system of linear equations with the coefficients $c_1, c_2, ..., c_n$ as unknowns. If the only solution to this system is the trivial solution (all coefficients are zero), then the vectors are linearly independent. If there are non-trivial solutions, the vectors are linearly dependent.

This method is particularly effective when dealing with vectors in $\mathbb{R}^n$ (n-dimensional Euclidean space), as it directly translates the vector equation into a familiar algebraic problem.

2. Using the Determinant (for n Vectors in $\mathbb{R}^n$)

This method is a shortcut that works specifically when you have n vectors in $\mathbb{R}^n$. You can form a matrix where each column (or row) is one of the vectors. Then, you calculate the determinant of this matrix. If the determinant is non-zero, the vectors are linearly independent. If the determinant is zero, they are linearly dependent.

Why does this work? The determinant of a matrix is related to the volume of the parallelepiped formed by the column vectors of the matrix. If the vectors are linearly dependent, the parallelepiped collapses into a lower dimension (e.g., a plane or a line), and its volume is zero. Conversely, if the vectors are linearly independent, the parallelepiped has a non-zero volume.

This method is computationally efficient, especially for smaller sets of vectors. However, it's crucial to remember that it only applies when the number of vectors equals the dimension of the space.

3. Row Reduction (Gaussian Elimination)

Similar to the determinant method, row reduction involves creating a matrix from the vectors and then performing row operations to reduce the matrix to its row-echelon form or reduced row-echelon form. The linear independence of the vectors can be determined by examining the resulting matrix.

If the reduced matrix has a pivot (leading 1) in every column, then the vectors are linearly independent. This is because each variable (coefficient) corresponds to a pivot, implying a unique solution (the trivial solution). If there is a column without a pivot, it indicates the presence of free variables, leading to non-trivial solutions and linear dependence.

Row reduction is a powerful technique that can be used even when the number of vectors is not equal to the dimension of the space. It provides a systematic way to solve the system of equations and determine the linear independence of the vectors.

For our problem, we'll stick with the system of equations method because it's the most fundamental and helps solidify the underlying concept. Let's jump into the solution!

Solving the Problem: System of Equations Method

Alright, let's put our knowledge into action! We need to show that the vectors $\vec{i}+2 \vec{j}+3 \vec{k}, 2 \vec{i}+\vec{j}+3 \vec{k}$, and $\vec{i}+\vec{j}+\vec{k}$ are linearly independent. As we discussed, we start with the equation:

c1(i⃗+2j⃗+3k⃗)+c2(2i⃗+j⃗+3k⃗)+c3(i⃗+j⃗+k⃗)=0⃗c_1(\vec{i}+2 \vec{j}+3 \vec{k}) + c_2(2 \vec{i}+\vec{j}+3 \vec{k}) + c_3(\vec{i}+\vec{j}+\vec{k}) = \vec{0}

where $c_1$, $c_2$, and $c_3$ are scalars. Our goal is to prove that the only solution to this equation is $c_1 = c_2 = c_3 = 0$.

Step 1: Expand and Group Components

Let's expand the equation and group the terms with the same unit vectors ($\vec{i}, \vec{j}, \vec{k}$):

(c1+2c2+c3)i⃗+(2c1+c2+c3)j⃗+(3c1+3c2+c3)k⃗=0i⃗+0j⃗+0k⃗(c_1 + 2c_2 + c_3)\vec{i} + (2c_1 + c_2 + c_3)\vec{j} + (3c_1 + 3c_2 + c_3)\vec{k} = 0\vec{i} + 0\vec{j} + 0\vec{k}

Step 2: Form the System of Equations

For this equation to hold, the coefficients of each unit vector must be equal. This gives us the following system of linear equations:

  1. c1+2c2+c3=0c_1 + 2c_2 + c_3 = 0

  2. 2c1+c2+c3=02c_1 + c_2 + c_3 = 0

  3. 3c1+3c2+c3=03c_1 + 3c_2 + c_3 = 0

Step 3: Solve the System of Equations

Now, we need to solve this system of equations. There are several ways to do this, such as substitution, elimination, or using matrices. Let's use the elimination method. We can eliminate $c_3$ from the first two equations by subtracting equation (2) from equation (1):

(c1+2c2+c3)βˆ’(2c1+c2+c3)=0βˆ’0(c_1 + 2c_2 + c_3) - (2c_1 + c_2 + c_3) = 0 - 0

This simplifies to:

βˆ’c1+c2=0-c_1 + c_2 = 0

So, we have:

  1. c2=c1c_2 = c_1

Now, let's eliminate $c_3$ from equations (1) and (3). We can do this by subtracting equation (3) from equation (1):

(c1+2c2+c3)βˆ’(3c1+3c2+c3)=0βˆ’0(c_1 + 2c_2 + c_3) - (3c_1 + 3c_2 + c_3) = 0 - 0

This simplifies to:

βˆ’2c1βˆ’c2=0-2c_1 - c_2 = 0

Substitute $c_2 = c_1$ (from equation 4) into this equation:

βˆ’2c1βˆ’c1=0-2c_1 - c_1 = 0

βˆ’3c1=0-3c_1 = 0

So,

  1. c1=0c_1 = 0

Since $c_2 = c_1$, we also have:

  1. c2=0c_2 = 0

Now, substitute $c_1 = 0$ and $c_2 = 0$ into equation (1):

0+2(0)+c3=00 + 2(0) + c_3 = 0

This gives us:

  1. c3=0c_3 = 0

Step 4: Conclusion

We have found that the only solution to the system of equations is $c_1 = 0$, $c_2 = 0$, and $c_3 = 0$. This is the trivial solution. Therefore, the vectors $\vec{i}+2 \vec{j}+3 \vec{k}, 2 \vec{i}+\vec{j}+3 \vec{k}$, and $\vec{i}+\vec{j}+\vec{k}$ are linearly independent.

Final Thoughts and Key Takeaways

Awesome! We've successfully proven that the given vectors are linearly independent. Let's recap the key steps and takeaways from this problem:

  • Understanding Linear Independence: Make sure you have a solid grasp of what linear independence means. It's the foundation for the entire process.
  • Setting up the Equation: The core idea is to start with the equation $c_1\vec{v}_1 + c_2\vec{v}_2 + ... + c_n\vec{v}_n = \vec{0}$ and try to prove that the only solution is the trivial one.
  • System of Equations: Expanding the equation and grouping components leads to a system of linear equations. This is where your algebra skills come in handy.
  • Solving the System: Use methods like substitution, elimination, or matrices to solve the system. The goal is to find the values of the coefficients $c_1, c_2, ..., c_n$.
  • Conclusion: If the only solution is the trivial solution (all coefficients are zero), the vectors are linearly independent. Otherwise, they are linearly dependent.

Linear independence is a fundamental concept in linear algebra, and understanding it will greatly help you in more advanced topics. Practice makes perfect, so try solving more problems like this one. You've got this!

If you have any questions or want to explore other methods, feel free to ask. Keep learning and keep exploring!