Problem 2
Question
Let $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right], \quad \mathbf{x}=\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right], \quad \text { and } \quad \mathbf{y}=\left[\begin{array}{l} y_{1} \\ y_{2} \end{array}\right] $$ (a) Show by direct calculation that \(A(\mathbf{x}+\mathbf{y})=A \mathbf{x}+A \mathbf{y}\). (b) Show by direct calculation that \(A(\lambda \mathbf{x})=\lambda(A \mathbf{x})\).
Step-by-Step Solution
Verified Answer
(a) \( A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y} \); (b) \( A(\lambda \mathbf{x}) = \lambda (A\mathbf{x}) \).
1Step 1: Add the Two Vectors
We are tasked with proving that the sum of the transformed vectors is equivalent to the transformation of the sum of the vectors. We begin by summing the vectors \( \mathbf{x} \) and \( \mathbf{y} \), resulting in: \[ \mathbf{x} + \mathbf{y} = \begin{bmatrix} x_1 + y_1 \ x_2 + y_2 \end{bmatrix}. \]
2Step 2: Multiply the Matrix by the Sum of Vectors
Next, multiply matrix \( A \) by the resulting vector sum: \[ A(\mathbf{x} + \mathbf{y}) = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} x_1 + y_1 \ x_2 + y_2 \end{bmatrix}.\] This gives us: \[ \begin{bmatrix} a_{11}(x_1 + y_1) + a_{12}(x_2 + y_2) \ a_{21}(x_1 + y_1) + a_{22}(x_2 + y_2) \end{bmatrix}. \]
3Step 3: Distribute Multiplication Over Addition
Distribute the matrix multiplication over the added vector components: \[ = \begin{bmatrix} a_{11}x_1 + a_{11}y_1 + a_{12}x_2 + a_{12}y_2 \ a_{21}x_1 + a_{21}y_1 + a_{22}x_2 + a_{22}y_2 \end{bmatrix}. \]
4Step 4: Separate Terms to Match Individual Vector Transformations
Rearrange and separate terms to match the effect of the matrix acting on the vectors individually: \[ = \begin{bmatrix} (a_{11}x_1 + a_{12}x_2) + (a_{11}y_1 + a_{12}y_2) \ (a_{21}x_1 + a_{22}x_2) + (a_{21}y_1 + a_{22}y_2) \end{bmatrix}. \]
5Step 5: Compute Matrix-Vector Product for Each Vector
Calculate \( A\mathbf{x} \) and \( A\mathbf{y} \) separately: \[ A\mathbf{x} = \begin{bmatrix} a_{11}x_1 + a_{12}x_2 \ a_{21}x_1 + a_{22}x_2 \end{bmatrix}, \quad A\mathbf{y} = \begin{bmatrix} a_{11}y_1 + a_{12}y_2 \ a_{21}y_1 + a_{22}y_2 \end{bmatrix}. \]
6Step 6: Add the Results of the Two Matrix-Vector Products
Sum the results from Step 5: \[ A\mathbf{x} + A\mathbf{y} = \begin{bmatrix} (a_{11}x_1 + a_{12}x_2) + (a_{11}y_1 + a_{12}y_2) \ (a_{21}x_1 + a_{22}x_2) + (a_{21}y_1 + a_{22}y_2) \end{bmatrix}. \]
7Step 7: Conclude Part (a)
This verifies our requirement: \( A(\mathbf{x} + \mathbf{y}) = A\mathbf{x} + A\mathbf{y} \).
8Step 8: Scale the Vector \(\mathbf{x}\) by \(\lambda\)
Now, for part (b), calculate \( \lambda \mathbf{x} = \begin{bmatrix} \lambda x_1 \ \lambda x_2 \end{bmatrix}. \)
9Step 9: Multiply the Matrix by the Scaled Vector
Multiply the matrix \( A \) by the scaled vector: \[ A(\lambda \mathbf{x}) = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \begin{bmatrix} \lambda x_1 \ \lambda x_2 \end{bmatrix}, \] resulting in \[ \begin{bmatrix} a_{11}(\lambda x_1) + a_{12}(\lambda x_2) \ a_{21}(\lambda x_1) + a_{22}(\lambda x_2) \end{bmatrix}. \]
10Step 10: Factor \(\lambda\) Out of the Result
Factor \( \lambda \) out from the expression: \[ = \lambda \begin{bmatrix} a_{11}x_1 + a_{12}x_2 \ a_{21}x_1 + a_{22}x_2 \end{bmatrix}. \]
11Step 11: Calculate \( \lambda (A \mathbf{x}) \) and Conclude
Observe that this expression is equivalent to: \( \lambda (A \mathbf{x}) \). Hence, \( A(\lambda \mathbf{x}) = \lambda (A \mathbf{x}) \).
Key Concepts
Matrix MultiplicationVector AdditionScalar MultiplicationDistributive Property
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra. It describes how we multiply a matrix by another matrix or a vector. It's not as straightforward as multiplying two numbers. When a matrix multiplies a vector, it transforms or alters the vector according to the rules encoded in the matrix. This operation involves combining the rows of the matrix with the columns of the vector, usually resulting in another vector.
In our exercise, we have:
In our exercise, we have:
- Matrix \( A = \begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{bmatrix} \)
- Vectors \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) and \( \mathbf{y} = \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \)
Vector Addition
Vector addition is another key operation in linear algebra. It is all about combining two vectors to form a new vector. In this process, each component of one vector is added to the corresponding component of another vector.
For instance, if we consider vectors \( \mathbf{x} \) and \( \mathbf{y} \):
For instance, if we consider vectors \( \mathbf{x} \) and \( \mathbf{y} \):
- \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \)
- \( \mathbf{y} = \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \)
Scalar Multiplication
Scalar multiplication involves multiplying a vector by a scalar (a single number), resulting in a vector that is scaled by that number. This operation affects every component of the vector equally.
Consider a scalar \( \lambda \) and vector \( \mathbf{x} \):
Consider a scalar \( \lambda \) and vector \( \mathbf{x} \):
- \( \lambda \in \mathbb{R} \) (real number)
- \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \)
Distributive Property
The distributive property is a fundamental algebraic principle allowing us to expand expressions in a way that simplifies calculations. It states that multiplying a sum by a number (or another algebraic entity) is the same as doing each multiplication separately and then adding the results. Mathematically, it says:\[ a(b + c) = ab + ac \]This property holds in both arithmetic and algebra, including for matrices and vector operations.
In our exercise, the distributive property allowed us to show:
In our exercise, the distributive property allowed us to show:
- How matrix \( A \) acting on the sum of vectors \( \mathbf{x} + \mathbf{y} \) produces the same result as \( A \) acting on each vector individually and then summing the results.
- For a scalar \( \lambda \) and vector \( \mathbf{x} \), how \( A(\lambda \mathbf{x}) \) simplifies to \( \lambda (A \mathbf{x}) \).
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