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Create a diagonal matrix \(D\) with the command \(D=\operatorname{diag}([1,2,3])\). It is easy to calculate powers of a diagonal matrix. Enter \(D^{\wedge} 2, D^{\wedge} 3, D^{\wedge} 4\), etc. Do you see a pattern emerging? If $$ D=\left[\begin{array}{rrrr} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{array}\right] $$ then what is \(D^{k}\) ?

Short Answer

Expert verified
\( D^k \) is diagonal with entries \( \lambda_i^k \) on the diagonal.

Step by step solution

01

Create the Diagonal Matrix

The diagonal matrix \( D \) is created using the command \( D = \operatorname{diag}([1, 2, 3]) \). This results in the following matrix: \[D = \begin{bmatrix} 1 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{bmatrix} \] The main diagonal consists of the elements 1, 2, and 3.
02

Calculate Squared Diagonal Matrix \(D^{\wedge 2}\)

To calculate \( D^2 \), each diagonal entry \( \lambda_i \) of \( D \) is squared. Thus, \( D^2 = \operatorname{diag}([1^2, 2^2, 3^2]) = \operatorname{diag}([1, 4, 9]) \). The resulting matrix is: \[D^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 4 & 0 \ 0 & 0 & 9 \end{bmatrix} \]
03

Calculate Cubed Diagonal Matrix \(D^{\wedge 3}\)

For \( D^3 \), cube each diagonal entry \( \lambda_i \). Therefore, \( D^3 = \operatorname{diag}([1^3, 2^3, 3^3]) = \operatorname{diag}([1, 8, 27]) \). The resulting matrix is: \[D^3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 8 & 0 \ 0 & 0 & 27 \end{bmatrix} \]
04

Calculate Quartic Diagonal Matrix \(D^{\wedge 4}\)

To calculate \( D^4 \), raise each entry \( \lambda_i \) to the fourth power. Thus, \( D^4 = \operatorname{diag}([1^4, 2^4, 3^4]) = \operatorname{diag}([1, 16, 81]) \). The resulting matrix is: \[D^4 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 16 & 0 \ 0 & 0 & 81 \end{bmatrix} \]
05

Identify Pattern in Powers of Diagonal Matrix

From the calculations in Steps 2 to 4, a pattern emerges: powering the diagonal matrix \( D \) involves raising each diagonal element \( \lambda_i \) to the power being considered, \( k \). Thus, \( D^k = \operatorname{diag}([\lambda_1^k, \lambda_2^k, \lambda_3^k]) \).
06

Final Step: General Expression for Powers of Diagonal Matrix

Using the observed pattern, for a general \( n \times n \) diagonal matrix \( D \) with diagonal entries \( \lambda_1, \lambda_2, \ldots, \lambda_n \), the matrix raised to the power \( k \) is given by: \[ D^k = \begin{bmatrix} \lambda_1^k & 0 & \cdots & 0 \ 0 & \lambda_2^k & \cdots & 0 \ \vdots & \vdots & \ddots & \vdots \ 0 & 0 & \cdots & \lambda_n^k \end{bmatrix} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Matrix Powers
Matrix powers are a fascinating area of study in linear algebra. When discussing the powers of a matrix, we're talking about multiplying a matrix by itself a certain number of times. For diagonal matrices, this concept becomes particularly simple. A diagonal matrix is characterized by elements on its main diagonal and zeros everywhere else.

When we raise a diagonal matrix to a power, we don't need to perform the traditional row-by-column multiplication. Instead, we can focus solely on the diagonal elements. If we have a diagonal matrix \( D \), with diagonal elements \( \lambda_1, \lambda_2,..., \lambda_n \), raising this matrix to the \( k \)th power involves raising each of these diagonal elements individually to the power of \( k \).
  • This is because the multiplication of zeros outside the diagonal has no contribution to the resulting matrix.
  • The structure of a diagonal matrix remains intact throughout, making calculations of powers straightforward.
Thus, if you have a diagonal matrix \( D \), calculating \( D^k \) simplifies to replacing each diagonal entry with its respective power: \( D^k = \operatorname{diag}([\lambda_1^k, \lambda_2^k, ..., \lambda_n^k]) \).

This unique property is both time-saving and computationally efficient, especially in programming and large-scale calculations.
Diagonal Elements
Diagonal elements are the heart of a diagonal matrix. They are the elements on the main diagonal of a matrix, where the row and column indices are equal. In a diagonal matrix, these are the only non-zero elements, making them vitally important. Each diagonal element \( \lambda_i \) has a significant role when performing matrix operations, especially matrix powers.

In the context of diagonal matrix \( D \):
  • The diagonal elements directly dictate the matrix's behavior.
  • They are powerful in determining the main properties like determinant and trace of the matrix.
  • For a square matrix, each \( \lambda_i \) can be viewed as an eigenvalue.
Consequently, when we raise a diagonal matrix to a power, each diagonal element is raised to that power. The zeros off the diagonal mean no additional values are introduced, keeping the operation as simple as powering numbers.

Understanding diagonal elements simplifies many matrix-related calculations and eases computational burdens when dealing with large datasets or matrices.
Matrix Operations
The world of matrix operations is vast, with efficiency often hinging on the type of matrix we are working with. Diagonal matrices are one of the simplest forms, allowing for streamlined operations versus general matrices.

Diagonal matrices simplify many operations:
  • **Addition**: Adding two diagonal matrices involves simply adding their corresponding diagonal elements.
  • **Multiplication**: When multiplying diagonal matrices, multiply the corresponding diagonal elements. Non-diagonal elements remain zero.
  • **Inversion**: For diagonal matrices that are not zero on the diagonal, inversion is straightforward: the reciprocal of each diagonal element forms the diagonal of the inverse matrix.
Efficient operations with diagonal matrices arise because they avoid the complex row-by-column multiplication required by general matrices, due to the presence of zeros outside the diagonal.

In programming contexts like MATLAB, operations with diagonal matrices can greatly reduce computational time, enabling quick processing of large amounts of data.
MATLAB Commands
MATLAB is a powerful tool for matrix calculations, offering pre-configured commands that make working with matrices easy and efficient.

The "diag" command in MATLAB is particularly handy for creating diagonal matrices. When we execute a command like \( D = \operatorname{diag}([1, 2, 3]) \), MATLAB instantly generates the diagonal matrix \( D \) with 1, 2, and 3 as its diagonal elements.
  • The "diag" function can construct matrices from vectors or extract diagonal elements from existing matrices.
  • It streamlines operations, particularly matrix powers, allowing users to focus on modifying the main diagonal alone.
Furthermore, matrix powers in MATLAB can be handled using the "D^k" notation. MATLAB computes these powers effectively by raising each diagonal element independently while maintaining zeroes off the diagonal.

Using MATLAB for matrix operations not only speeds up calculations but also provides a clear and straightforward way to manipulate and analyze matrices of various types and sizes.

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Most popular questions from this chapter

For the matrices in Exercises \(13-19\), find the eigenvalues and eigenvectors. You may use any method you wish as long as you choose eigenvectors with all integer entries. For each eigenvalue state the algebraic and the geometric multiplicity. $$ \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] $$

For the matrices in Exercises \(13-19\), find the eigenvalues and eigenvectors. You may use any method you wish as long as you choose eigenvectors with all integer entries. For each eigenvalue state the algebraic and the geometric multiplicity. $$ \left[\begin{array}{rrr} -2 & 2 & -1 \\ -6 & 5 & -2 \\ 4 & -2 & 3 \end{array}\right] $$

In Exercises \(78-88\) find a fundamental set of solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\). Solve the initial value problem with \(\mathbf{x}(0)=\mathbf{x}_{0}\). $$ A=\left[\begin{array}{rrrr} -8 & -7 & 10 & -2 \\ 7 & 6 & -12 & 2 \\ 1 & 1 & -3 & 0 \\ 5 & 3 & -2 & -1 \end{array}\right], \quad \mathbf{x}_{0}=\left[\begin{array}{r} -2 \\ -8 \\ -6 \\ 1 \end{array}\right] $$

In Exercises \(78-88\) find a fundamental set of solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\). Solve the initial value problem with \(\mathbf{x}(0)=\mathbf{x}_{0}\). $$ A=\left[\begin{array}{rrrr} -9 & 7 & -27 & 17 \\ 7 & -4 & 18 & -13 \\ 4 & -1 & 8 & -7 \\ -1 & 4 & -9 & 3 \end{array}\right], \quad \mathbf{x}_{0}=\left[\begin{array}{l} 5 \\ 0 \\ 0 \\ 2 \end{array}\right] $$

Given that \(\lambda=-1\) is an eigenvalue of the matrix $$ A=\left[\begin{array}{rrr} 3 & 6 & 10 \\ 0 & -1 & 0 \\ 0 & -1 & -2 \end{array}\right] $$ reduce the augmented matrix \(M=[A-(-1) *\) eye \((3)\), zeros \((3,1)]\) and interpret the result to find the associated eigenvector. Use null \(\left(A-(-1) *\right.\) eye \(\left.(3), ' r^{\prime}\right)\) and compare results. Perform similar analysis for the remaining eigenvalues, \(\lambda=-2\) and \(\lambda=3\).

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