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Magazine Chapter 8: Problem 26
If \(Z\) is an idempotent matrix, th?n \((I+Z)^{n}\) a. \(I+2^{n} Z\) b. \(I+\left(2^{n}-1\right) Z\) c. \(I-\left(2^{\prime \prime}-1\right) Z\) d. none of these
Short Answer
Expert verified
The correct solution is (b) \(I + (2^n - 1) Z\).
Step by step solution
01
Understand the properties of the idempotent matrix
A matrix \(Z\) is idempotent if \(Z^2 = Z\). This property plays a crucial role in the simplification of expressions involving \(Z\).
02
Recall the Binomial Theorem
The Binomial Theorem states that \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\)., where \(\binom{n}{k}\) is the binomial coefficient. In this exercise, let \(x = I\) (identity matrix) and \(y = Z\).
03
Expand \((I+Z)^n\) using the Binomial Theorem
Applying the Binomial Theorem, we find: \((I+Z)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} Z^k\). Since \(I^k = I\) for any positive integer \(k\), this reduces to \((I+Z)^n = \sum_{k=0}^{n} \binom{n}{k} I Z^k\).
04
Utilize the properties of the idempotent matrix in the expansion
Given that \(Z\) is idempotent, for \(k \geq 2\), we know \(Z^k = Z\). This means the expansion becomes \((I+Z)^n = I + nZ + \binom{n}{2}Z + \cdots \).
05
Simplify the series
Notice that, because each term in the series with \(Z\) is also equal to \(Z\), the expression simplifies greatly. The sum \(\sum_{k=1}^{n} \binom{n}{k} Z\) will result in \(Z(\sum_{k=1}^{n} \binom{n}{k})\) which simplifies to \(Z(2^n - 1)\).
06
Final expression
Combine the identity component with the factorized \(Z\) term. Thus, the expression simplifies to \(I + (2^n - 1)Z\). This means the correct answer is option b: \(I + (2^n - 1) Z\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Matrix Properties
A matrix is a collection of numbers arranged into a fixed number of rows and columns. It is a fundamental concept in linear algebra and has various important properties that are useful in different applications.
One specific type of matrix you might encounter is an **idempotent matrix**. This type of matrix satisfies the condition where multiplying the matrix by itself results in the same matrix. Mathematically, if a matrix \( Z \) is idempotent, then \( Z^2 = Z \). This means if you apply any power greater than or equal to 2, such as \( Z^3, Z^4 \), etc., the result is still \( Z \).
One specific type of matrix you might encounter is an **idempotent matrix**. This type of matrix satisfies the condition where multiplying the matrix by itself results in the same matrix. Mathematically, if a matrix \( Z \) is idempotent, then \( Z^2 = Z \). This means if you apply any power greater than or equal to 2, such as \( Z^3, Z^4 \), etc., the result is still \( Z \).
- Idempotent matrices are not just mathematical oddities; they have practical applications in statistics, especially in linear regression where they help create the projection matrix which projects observed vectors onto a subspace.
- The unique property \( Z^2 = Z \) reduces the complexity of many calculations involving idempotent matrices.
Binomial Theorem Application
The **Binomial Theorem** provides a way to expand expressions raised to a power. It states that for any numbers \( x \) and \( y \), and a non-negative integer \( n \), the expression \((x+y)^n\) can be expanded into a sum involving terms of the form \( \binom{n}{k} x^{n-k} y^k \).
This theorem is particularly useful when working with matrices, as it allows us to systematically expand and simplify expressions. In our exercise, we used the identity matrix \( I \) and the idempotent matrix \( Z \) in the context of \((I + Z)^n\).
This theorem is particularly useful when working with matrices, as it allows us to systematically expand and simplify expressions. In our exercise, we used the identity matrix \( I \) and the idempotent matrix \( Z \) in the context of \((I + Z)^n\).
- By using the binomial theorem, \((I+Z)^n\) is expressed as a series: \[ \sum_{k=0}^{n} \binom{n}{k} I^{n-k} Z^{k} \].
- This expansion helps us leverage the unique identities of \( I \) and \( Z \) to find a simplified form of the overall matrix expression.
Identity Matrix
The **identity matrix** is a special type of matrix that plays a major role in linear algebra. It's like the number 1 in matrix operations similar to how multiplying any number by 1 leaves the number unchanged.
For an identity matrix \( I \) of size \( n \times n \), it has 1s on the main diagonal (from the top left to the bottom right) and 0s elsewhere. The identity matrix has the property that for any matrix \( A \) of the same size, \( I A = A I = A \).
For an identity matrix \( I \) of size \( n \times n \), it has 1s on the main diagonal (from the top left to the bottom right) and 0s elsewhere. The identity matrix has the property that for any matrix \( A \) of the same size, \( I A = A I = A \).
- When using the identity matrix in the binomial theorem context, \( I^k = I \) for any non-negative power \( k \), because the identity matrix multiplied by itself any number of times remains unchanged.
- This property simplifies many expressions during matrix operations, making calculations involving higher powers much more manageable.
Binomial Coefficients
**Binomial coefficients** are essential components of the binomial theorem. They describe the number of ways you can choose \( k \) items from \( n \) items without regarding the order. This is mathematically denoted as \( \binom{n}{k} \) and called a "choose" function.
In our exercise, they appear in the expansion \((I+Z)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} Z^k\).
In our exercise, they appear in the expansion \((I+Z)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} Z^k\).
- Binomial coefficients are calculated using the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \(!\) denotes factorial.
- These coefficients help in distributing powers across the terms \( I \) and \( Z \) while maintaining the order and combination that forms the binomial expansion.
