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Chapter 5: Problem 20

For \(A=\left[\begin{array}{ll}0 & 2 \\ 0 & 5\end{array}\right]\), find the powers \(A^{k}\) (including \(A^{0}\) ) and show explicitly that their sum agrees with \((I-A)^{-1}\).

Short Answer

Expert verified
The sum \(S\) and \((I-A)^{-1}\) do not agree because of series non-convergence.

Step by step solution

01

Identify the Base Case

We start with the definition: Any matrix raised to the power of zero, \(A^0\), equals the identity matrix. For a \(2 \times 2\) matrix, the identity matrix \(I\) will be \(\left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right]\). Therefore, \(A^0 = I\).
02

Calculate the First Power

Compute \(A^1\), which is simply \(A\) itself: \(A = \left[ \begin{array}{cc} 0 & 2 \ 0 & 5 \end{array} \right]\).
03

Calculate the Second Power

To find \(A^2\), compute \(A \times A\):\[A^2 = \left[ \begin{array}{cc} 0 & 2 \ 0 & 5 \end{array} \right] \times \left[ \begin{array}{cc} 0 & 2 \ 0 & 5 \end{array} \right] = \left[ \begin{array}{cc} 0 & 10 \ 0 & 25 \end{array} \right]\].
04

General Formula for \(A^k\)

Observe a pattern for the powers of \(A\). If we continue this pattern:\[A^3 = \left[ \begin{array}{cc} 0 & 50 \ 0 & 125 \end{array} \right], \ldots \]Notice the pattern for any power:\[A^k = \left[ \begin{array}{cc} 0 & 2^k \ 0 & 5^k \end{array} \right]\].
05

Compute the Sum \(S = I + A + A^2 + \ldots\)

The sum of the infinite series is:\[S = I + A + A^2 + A^3 + \ldots\]Substitute the identity matrix and the pattern:\[S = \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] + \left[ \begin{array}{cc} 0 & 2 \ 0 & 5 \end{array} \right] + \left[ \begin{array}{cc} 0 & 10 \ 0 & 25 \end{array} \right] + \ldots\]Each element involves a geometric series.
06

Simplify Each Element of the Sum

For the geometric series in each element:- Top row: Constant 0, so sum remains 0.- Bottom row diagonal is a series: \(1 + 5 + 5^2 + \ldots = \frac{1}{1-5}\) (since it's a geometric series with common ratio 5).- Top-right term: \(2 + 2^2 + 2^3 + \ldots = \frac{2}{1-2}\) is not convergent because it's non-geometric for series with ratio \(>|1|\).Sum cannot be accurately described under convergence issues, highlighting necessary checks for each component.
07

Compute \((I-A)^{-1}\)

Calculate the inverse of \((I - A)\):Let \(B = I - A\):\[B = \left[ \begin{array}{cc} 1 & 0 \ 0 & 1 \end{array} \right] - \left[ \begin{array}{cc} 0 & 2 \ 0 & 5 \end{array} \right] = \left[ \begin{array}{cc} 1 & -2 \ 0 & -4 \end{array} \right]\].Compute \((I - A)^{-1}\):\[B^{-1} = \left[ \begin{array}{cc} 1 & 2/4 \ 0 & -1/4 \end{array} \right]\].
08

Conclusion and Agreement Check

The sum \(S = I + A + A^2 + \ldots\) does not agree with \((I-A)^{-1}\) due to geometric series non-convergence. The geometric sum and matrix inverse do not align due to a non-converging series aspect in matrix operations for this problem.

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Key Concepts

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

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant, known as the common ratio. For example, the series 1, 2, 4, 8, ... doubles each term from the one before, giving it a common ratio of 2.

The sum of a geometric series can be calculated by the formula: \[ S = rac{a}{1 - r} \] where \( a \) is the first term and \( r \) is the common ratio. This formula applies only if the series converges, meaning the absolute value of \( r < 1 \).

In the context of matrices, each element of a matrix can form its own geometric series when computing powers of the matrix. However, convergence is still key. If part of the series does not converge, like here with a series involving \( 2 \) where \( |2| > 1 \), it means you can't rely on a simple sum formula for that term.
Matrix Inversion
Matrix inversion is a process similar to finding the reciprocal of a number but for a square matrix. When we say "invert a matrix," we signify finding another matrix that, when multiplied by the original, results in the identity matrix. The identity matrix acts like the number 1 in matrix multiplication.

However, not every matrix has an inverse. A matrix is invertible if it is square and has a non-zero determinant. If matrix \( A \) has an inverse, it is denoted as \( A^{-1} \) and satisfies \( AA^{-1} = A^{-1}A = I \).

For the problem exercise, we were tasked to find \((I - A)^{-1}\). First, we find \( B = I - A \), and then we find \( B^{-1} \). Inverting matrices involves calculations like finding determinants and adjugates, which can be complex but are necessary for solutions involving matrix powers.
Convergence in Matrices
Convergence in matrices refers to the behavior when raising a matrix to increasing powers and summing those results. Not all sequences of matrix powers will converge, similar to real number series.

For convergence in matrix operations:
  • Each element of the matrix powers should determine a convergent behavior.
  • The array position equivalent geometric series should converge for the overall matrix power series to make sense.
In the problem provided, the sum \( S = I + A + A^2 + \ldots \) was aimed to be equated to \((I-A)^{-1}\). However, convergence problems arise due to elements having geometric series with ratios \( |r| > 1 \), notably like 2 in our example.

Understanding convergence in matrices helps recognize valid solutions and avoid errors in computations involving matrix power series.

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Most popular questions from this chapter

With a friction matrix \(F\) in the equation \(u^{\prime \prime}+F u^{\prime}-A u=0\), substitute a pure exponential \(u=e^{\lambda t} x\) and find a quadratic eigenvalue problem for \(\lambda\).

Prove in three steps that \(A^{\mathrm{T}}\) is always similar to \(A\) (we know that the \(\lambda\) 's are the same, the eigenvectors are the problem): (a) For \(A=\) one block, find \(M_{i}=\) permutation so that \(M_{i}^{-1} J_{i} M_{i}=J_{i}^{\mathrm{T}}\), (b) For \(A=\) any \(J\), build \(M_{0}\) from blocks so that \(M_{0}^{-1} J M_{0}=J^{\mathrm{T}}\). (c) For any \(A=M J M^{-1}\) : Show that \(A^{\mathrm{T}}\) is similar to \(J^{\mathrm{T}}\) and so to \(J\) and to \(A\).

If \(J\) is the 5 by 5 Jordan block with \(\lambda=0\), find \(J^{2}\) and count its eigenvectors, and find its Jordan form (two blocks).

Which of these matrices \(A_{1}\) to \(A_{6}\) are similar? Check their eigenvalues. $$ \left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \quad\left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 0 \\ 1 & 1 \end{array}\right] \quad\left[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right] $$

Suppose there are three major centers for Move-It-Yourself trucks. Every month half of those in Boston and in Los Angeles go to Chicago, the other half stay where they are, and the trucks in Chicago are split equally between Boston and Los Angeles. Set up the 3 by 3 transition matrix \(A\), and find the steady state \(u_{\infty}\) corresponding to the eigenvalue \(\lambda=1\).
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