Problem 42 - Differential Equations and Linear Algebra

Question

Consider the vectors $$ A_{1}=\left[\begin{array}{rl} 1 & 2 \\ -1 & 3 \end{array}\right], A_{2}=\left[\begin{array}{rr} -2 & 1 \\ 1 & -1 \end{array}\right] $$ in \(M_{2}(\mathbb{R}) .\) Find \(\operatorname{span}\left\\{A_{1}, A_{2}\right\\},\) and determine whether or not \(B=\left[\begin{array}{rl}3 & 1 \\\ -2 & 4\end{array}\right]\) lies in this subspace.

Step-by-Step Solution

Answer

Matrix \(B\) is in the span of \(A_1\) and \(A_2\), as it can be represented as a linear combination of these two matrices using the scalar coefficients \(k_1 = 1\) and \(k_2 = -1\), i.e., \(B = 1 A_1 + (-1) A_2\).

Step 1: Represent B as a linear combination of A1 and A2

Let's represent \(B\) as a linear combination of \(A_1\) and \(A_2\) : \(B = k_1 A_1 + k_2 A_2\) Where \(k_1\) and \(k_2\) are scalar coefficients.

Step 2: Set up equations

We can write the above equation in terms of matrix elements, \(\left[\begin{array}{rl} 3 & 1 \\ -2 & 4 \end{array}\right] = k_1 \left[\begin{array}{rl} 1 & 2 \\ -1 & 3 \end{array}\right] + k_2 \left[\begin{array}{rr} -2 & 1 \\ 1 & -1 \end{array}\right]\) Now, we can set up a system of equations by matching the corresponding elements of the matrices: \(3 = k_1(1) + k_2(-2)\) (1) \(1 = k_1(2) + k_2(1)\) (2) \(-2 = k_1(-1) + k_2(1)\) (3) \(4 = k_1(3) + k_2(-1)\) (4)

Step 3: Solve the system of equations

We can now solve this system of linear equations for \(k_1\) and \(k_2\). Let's first solve equations (1) and (2): From equation (1), we get: \(k_2 = \dfrac{k_1 - 3}{2}\) Now substitute this value of \(k_2\) in equation (2): \(1 = 2k_1 + \dfrac{k_1 - 3}{2}\) \(2 = 4k_1 + (k_1 - 3)\) \(5 = 5k_1\) \(k_1 = 1\) Now we can use the value of \(k_1\) to find the value of \(k_2\): \(k_2 = \dfrac{1 - 3}{2} = -1\)

Step 4: Form the linear combination and conclusion

Now that we have the values of \(k_1\) and \(k_2\), let's rewrite the linear combination of \(A_1\) and \(A_2\): \(B = 1 A_1 + (-1) A_2\) Thus, matrix \(B\) is in the span of \(A_1\) and \(A_2\), as it can be represented as a linear combination of these two matrices using the scalar coefficients \(k_1 = 1\) and \(k_2 = -1\).

Key Concepts

Linear Combination

A linear combination involves creating a new vector or matrix by summing together scalar multiples of existing vectors or matrices. This concept lets you express one vector as a mix of other vectors. Think of it like creating a new paint color by mixing different amounts of primary colors.
For a vector or matrix to be a linear combination of others, you can form equations comparing elements. For example, with the matrices \(A_1\) and \(A_2\) given in the problem, matrix \(B\) is a linear combination if you can find scalars \(k_1\) and \(k_2\) such that:

• \(B = k_1 A_1 + k_2 A_2\)

The task involves setting up equations for each element and solving for these scalars, much like determining the recipe for a new dish using available ingredients.

Span of Vectors

The span of a set of vectors (or matrices) is a crucial concept that captures all possible linear combinations. It represents a subspace containing all vectors you can construct using these combinations. In simple terms, it shows the space covered by stretching and combining the given vectors in all different proportions.
In our example, you're asked to find the span of \(\{A_1, A_2\}\). It includes every matrix you can make using any linear combination of \(A_1\) and \(A_2\). Notably, the span forms a subspace in the matrix space \(M_2(\mathbb{R})\), meaning any matrix in this subspace can be written using \(A_1\) and \(A_2\) through appropriate scalar multiplication and addition.
Determining if matrix \(B\) lies in this span requires showing \(B\) can be constructed from \(A_1\) and \(A_2\), underscoring how span highlights the foundational building blocks of a vector space.

System of Linear Equations

A system of linear equations consists of multiple equations involving the same set of variables. These systems are essential for solving problems like determining the coefficients in a linear combination. Typically shown as matrices or equations in the form \(ax + by = c\), they help in finding values that satisfy all given conditions simultaneously.
In the context of our exercise, representing \(B\) as a linear combination requires setting up such a system. By equating corresponding elements from \(B\) to the combination of \(A_1\) and \(A_2\), we create a series of equations:

• 3 = \(k_1\)(1) + \(k_2\)(-2)
• 1 = \(k_1\)(2) + \(k_2\)(1)
• -2 = \(k_1\)(-1) + \(k_2\)(1)
• 4 = \(k_1\)(3) + \(k_2\)(-1)

Solving these equations using substitution or elimination reveals the necessary scalars \(k_1\) and \(k_2\), illustrating how tightly systems of equations complement linear combinations and spans in understanding vector spaces.