Question

The given matrix \(A\) is diagonalizable. (a) Find \(T\) and \(D\) such that \(T^{-1} A T=D\). (b) Using (12c), determine the exponential matrix \(e^{A t}\).\(A=\left[\begin{array}{ll}5 & -6 \\ 3 & -4\end{array}\right]\)

Short answer

Answer: The matrix exponential e^{At} is: $e^{At} = \begin{pmatrix} e^t \cos(t) & -2e^t \sin(t) \\ 2e^t \sin(t) & e^t \cos(t) \end{pmatrix}$

Step-by-step solution

Find Eigenvalues of A

First, let's find the eigenvalues of the given matrix A. To do this, we set up the characteristic equation: |A - λI| = 0 where λ represents the eigenvalues. In our case, $\begin{vmatrix} 5 - λ & -6\\ 3 & -4 - λ \end{vmatrix} = (5 - λ)(-4 - λ) - (-6)(3) = 0$ Calculating this yields: \((5 - \lambda)(-4 - \lambda) + 18 = \lambda^2 - \lambda + 2 = 0\) The eigenvalues are λ1 = 1 + i and λ2 = 1 - i.

Find Eigenvectors Corresponding to Each Eigenvalue

Now, we need to find the eigenvectors corresponding to each eigenvalue. We will solve the following system for each eigenvalue: (A - λI)X = 0 For λ1 = 1 + i: $\begin{pmatrix} 4 - i & -6\\ 3 & -3 - i \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ Solving this system, we get the eigenvector corresponding to λ1 is: $X_1 = \begin{pmatrix} 2i \\ 1+ i \end{pmatrix}$ For λ2 = 1 - i: $\begin{pmatrix} 4 + i & -6\\ 3 & -3 + i \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ Solving this system, we get the eigenvector corresponding to λ2 is: $X_2 = \begin{pmatrix} -2i \\ 1- i \end{pmatrix}$

Form T and Find T^{-1}

Now, let's form the matrix T using the eigenvectors we found: $T = \begin{pmatrix} 2i & -2i \\ 1 + i & 1 - i \end{pmatrix}$ Next, we find T^{-1}: $T^{-1} = \dfrac{1}{(2i)(1-i)-(1+i)(-2i)} \begin{pmatrix} 1 - i & 2i \\ -i - 1 & 2i \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 1 - i & 2i \\ -i - 1 & 2i \end{pmatrix}$

Calculate Diagonal Matrix D

Now, let's calculate the diagonal matrix D by evaluating T^{-1} A T: $D = T^{-1} A T = \frac{1}{4} \begin{pmatrix} 1 - i & 2i \\ -i - 1 & 2i \end{pmatrix} \begin{pmatrix} 5 & -6 \\ 3 & -4 \end{pmatrix} \begin{pmatrix} 2i & -2i \\ 1 + i & 1 - i \end{pmatrix}$ $D = \begin{pmatrix} 1+i & 0 \\ 0 & 1-i \end{pmatrix}$

Determine the Matrix Exponential e^{At}

Finally, determine the matrix exponential e^{At} using the equation: \(e^{At} = T e^{D t} T^{-1}\) $e^{D t} = \begin{pmatrix} e^{(1+i)t} & 0 \\ 0 & e^{(1-i)t} \end{pmatrix}$ So, $e^{At} = \begin{pmatrix} 2i & -2i \\ 1 + i & 1 - i \end{pmatrix} \begin{pmatrix} e^{(1+i)t} & 0 \\ 0 & e^{(1-i)t} \end{pmatrix} \frac{1}{4}\begin{pmatrix} 1 - i & 2i \\ -i - 1 & 2i \end{pmatrix}$ After the calculation, we get: $e^{At} = \begin{pmatrix} e^t \cos(t) & -2e^t \sin(t) \\ 2e^t \sin(t) & e^t \cos(t) \end{pmatrix}$

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial for working with matrices and their applications. Simply put, eigenvalues are special numbers associated with a square matrix that provide insights into the matrix's properties. They are found by solving the characteristic equation, which is formed by subtracting a scalar multiple of the identity matrix from the original matrix and setting the determinant to zero:

\[ \text{det}(A - \text{λ}I) = 0 \]
In the context of the exercise, the eigenvalues \( \text{λ}_1 = 1 + i \) and \( \text{λ}_2 = 1 - i \) represent the 'scaling factors' that indicate how the matrix 'A' stretches or compresses its eigenvectors.

Eigenvectors, on the other hand, are non-zero vectors that change at most by their scalar multiple after the matrix transformation. They are found by plugging each eigenvalue back into the equation \[ (A - \text{λ}I)\text{X} = 0 \] and solving for the vector \( \text{X} \). The eigenvectors corresponding to our calculated eigenvalues are \( \text{X}_1 \) and \( \text{X}_2 \) from the solution steps. These concepts lay the groundwork for many advanced topics including matrix diagonalization, stability analysis, and more.

Matrix Exponentials

The matrix exponential is a vital concept in solving systems of differential equations, especially linear ones. Matrix exponentiation involves raising the special number 'e' (the base of the natural logarithm) to the power of a matrix, typically seen as \( e^{At} \), where 'A' is a matrix and 't' is a real number. In the given exercise, the matrix exponential is used to find the exponential of matrix 'A'.

For a diagonalizable matrix 'A', the matrix exponential can be computed using the formula \[ e^{At} = T e^{Dt} T^{-1} \], where 'T' is the matrix of eigenvectors and 'D' is the diagonal matrix of eigenvalues. This is much simpler than trying to calculate \( e^{At} \) directly. In practice, the exponent of a diagonal matrix 'D' is straightforward, as it only requires taking the exponential of each of the diagonal elements. The provided solution steps show this process, leading to a powerful way to solve systems of linear differential equations.

Systems of Differential Equations

Systems of differential equations are sets of equations that describe the dynamics of systems with multiple interrelated variables. These systems often arise in natural sciences and engineering when modeling complex phenomena. A key tool for solving linear systems of ordinary differential equations (ODEs) is diagonalization.

Diagonalization simplifies the system by decoupling the variables, allowing the equations to be solved individually. Once a matrix 'A' is diagonalizable (like in our exercise), one can use the eigenvalues and eigenvectors to transform the system into an easier one to manage. The key is that, with a diagonal matrix, the exponential function of the matrix can be computed much more straightforwardly. This becomes highly relevant in applications such as physics and finance where predicting behavior over time is crucial.

In the context of the original exercise, the student is effectively learning how to solve a linear system of ODEs using the technique of diagonalization. The final step where the matrix exponential \( e^{At} \) is computed represents the solution to a system of ODEs in the form \( \frac{d\text{y}}{dt} = A\text{y} \) for an initial value problem with the initial condition given by the identity matrix. The approach outlined certainly provides a clear link between algebraic concepts and their applications in calculus and differential equations.