Problem 17 - Differential Equations and Linear Algebra

Question

Determine whether the given set of vectors is linearly independent in \(M_{2}(\mathbb{R})\). $$A_{1}=\left[\begin{array}{rr} 2 & -1 \\ 3 & 4 \end{array}\right], A_{2}=\left[\begin{array}{rr} -1 & 2 \\ 1 & 3 \end{array}\right]$$.

Step-by-Step Solution

Answer

The given set of vectors (matrices) is linearly independent in \(M_2(\mathbb{R})\) as the only solution for the linear combination \(c_1 A_1 + c_2 A_2 = 0\) is the trivial solution, where \(c_1 = 0\) and \(c_2 = 0\).

Step 1: Define the linear combination

Let's set up a linear combination of the given matrices with scalar coefficients: \(c_1 A_1 + c_2 A_2 = 0\) where \(c_1\) and \(c_2\) are scalar values and 0 is the 2x2 zero matrix.

Step 2: Write the linear combination in matrix form

Let's write down the linear combination in matrix form: \(c_1 \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} + c_2 \begin{bmatrix} -1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Step 3: Expand the linear combination

Now, let's expand the linear combination: \(\begin{bmatrix} 2c_1 - c_2 & -c_1 + 2c_2 \\ 3c_1 + c_2 & 4c_1 + 3c_2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)

Step 4: Set up a system of linear equations

From the expanded linear combination, we can write down a system of linear equations by equating the corresponding elements: 1. \(2c_1 - c_2 = 0\) 2. \(-c_1 + 2c_2 = 0\) 3. \(3c_1 + c_2 = 0\) 4. \(4c_1 + 3c_2 = 0\) Notice that equations 1 and 3 are proportional, and equations 2 and 4 are proportional. So, we can eliminate one of the proportional equations (for example, equations 3 and 4) and keep only equations 1 and 2: 1. \(2c_1 - c_2 = 0\) 2. \(-c_1 + 2c_2 = 0\)

Step 5: Solve the system of linear equations

Now, let's solve the remaining system of linear equations: From equation 1, we have \(c_2 = 2c_1\). Substituting this into equation 2: \(-c_1 + 2(2c_1) = 0\) Simplifying the equation gives us: \(3c_1 = 0\) From this equation, we see that \(c_1 = 0\). Substituting the value of \(c_1\) back into the expression for \(c_2\), we obtain: \(c_2 = 2(0) = 0\)

Step 6: Conclude if the set is linearly independent or not

Since we found that the only solution to our system of linear equations is the trivial one, where \(c_1 = 0\) and \(c_2 = 0\), we can conclude that the set of vectors (matrices) is linearly independent in M_2(ℝ).

Key Concepts

Systems of Linear Equations

In mathematics, a system of linear equations involves finding solutions where multiple linear relationships coexist between the variables. For instance, linear equations typically appear in a form such as \( ax + by = c \). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) represent the variables.
When dealing with systems, multiple such equations are solved simultaneously. In the context of our exercise, matrix elements create a system by equating corresponding entries of matrix expressions to zero, forming equations like:

• \(2c_1 - c_2 = 0\)
• \(-c_1 + 2c_2 = 0\)

These equations are used to determine if the only solution is the trivial one, meaning that the coefficients \(c_1\) and \(c_2\) must both be zero. If this is the case, the matrices constitute a linearly independent set.
Linear independence signifies that none of the vector combinations can produce a zero vector except when all coefficients are zero. Hence, solving these equations helps us decide the linear independence of the matrices involved.

Matrix Algebra

Matrix algebra involves manipulating matrices through operations like addition, subtraction, multiplication, and finding determinants or inverses. Matrices provide a structured way to represent and solve systems of equations, and they display different properties compared to ordinary numbers.
When two matrices are equated, such as in linear combinations, each corresponding entry in the equation must be equal. This is why matrix operations become a powerful tool to translate multiple equations into a single matrix form.
In the given exercise, matrix algebra helps us transform the entire set of linear equations into a unified form:
\[ c_1 \begin{bmatrix} 2 & -1 \ 3 & 4 \end{bmatrix} + c_2 \begin{bmatrix} -1 & 2 \ 1 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \ 0 & 0 \end{bmatrix}\] By carrying out these operations, we created a bridge between each matrix equation's rows and columns. Matrix algebra simplifies the process of finding solutions by treating defined systems collectively, harnessing this representation to analyze problems with precision.

Vector Spaces

Vector spaces form the foundation of linear algebra, introducing structures where vectors can be added, subtracted, and multiplied (by scalars). A vector space is essentially a collection of vectors that maintains the rules of vector addition and scalar multiplication.
The relevance of vector spaces in our exercise lies in the concept of linear independence. A set of vectors (or matrices) is considered linearly independent if no vector can be represented as a complex combination of others using non-zero scalars. Each vector adds a new dimension to the space, allowing it to cover a broader area.
Since the matrices \(A_1\) and \(A_2\) in the exercise form a vector space in \(M_{2}(\mathbb{R})\), analyzing their linear independence tests whether these vectors can span a plane independently. If solely the trivial combination (\(c_1 = 0\), \(c_2 = 0\)) holds true, then they are deemed independent, contributing uniquely to the structure of the space without redundancy.
Understanding vector spaces allows us to visualize and apprehend the geometry and dimensionality inherent in complex algebraic concepts.